1 Gram (g) is equal to 0.001 kilogram (kg). To convert grams to kg, multiply the gram value by 0.001 or divide by 1000. For example, to convert 100 grams to kg, multiply 100 by 0.001, that makes 0.1 kg is 100g. Grams to kilograms formula. Kilogram = gram. 0.001. Kilogram = gram / 1000. 1 Gram = 0.001 Kilogram. Gram is a metric. Do I buy juice at the sale price of 2 bottles with 32 oz each for $6.00, or do I buy 1 bottle containing 72 oz for $6.99? Now you can use our unit price calculator to calculate the cost per ounce of one or both deals, and quickly figure out which is the better bargain. 1 US tsp (4.93ml). 3 = 1 US tbsp (14.79ml) Be careful, though - staying in the US doesn't exactly mean that you'll find only US teaspoons there. For example, in nutritional labeling and medicine, the teaspoon always has 5 ml, which is simply the metric teaspoon. Free online density converter - converts between 42 units of density, including kilogram/cubic meter, gram/cubic centimeter, kilogram/cubic centimeter, gram/cubic meter g/m^3, etc. Also, explore many other unit converters or learn more about density unit conversions.
Combustion Analysis
Ten Examples
This technique requires that you burn a sample of the unknown substance in a large excess of oxygen gas. The combustion products will be trapped separately from each other and the weight of each combustion product will be determined. From this, you will be able to calculate the empirical formula of the substance. This technique has been most often applied to organic compounds.
A brief discussion (yet quite informative) of the history of this technique can be found here. This technique is also called 'elemental analysis'
Some points to make about combustion analysis:
1) The elements making up the unknown substance almost always include carbon and hydrogen. Oxygen is often involved and nitrogen is involved sometimes. Other elements can be involved, but problems with C and H tend to predominate followed by C, H and O and then by C, H, O and N.
2) We must know the mass of the unknown substance before burning it.
3) All the carbon in the sample winds up as CO2 and all the hydrogen in the sample winds up as H2O.
4) If oxygen is part of the unknown compound, then its oxygen winds up incorporated into the oxides. The mass of oxygen in the sample will almost always be determined by subtraction.
5) Often the N is determined via a second experiment and this introduces a bit of complexity to the problem. Nitrogen dioxide is the usual product when nitrogen is involved. Sometimes the nitrogen product is N2, sometimes NH3.
6) Sometimes the problem asks you for the empirical formula and sometimes for the molecular formula (or both). Two points: (a) you have to know the molar mass to get to the molecular formula and (b) you have to calculate the empirical formla first, even if the question doesn't ask for it. A few lines below is a link that goes to a file that discusses how to go from empirical to molecular formulas.
Here is a brief overview of the solution steps before doing some examples:
1) Determine the grams of each element present in the original compound. Carbon is always in CO2 in the ratio (12.011 g / 44.0098 g), hydrogen is always in H2O in the ratio (2.0158 g / 18.0152 g), etc.
2) Convert grams of each elment to the number of moles. You do this by dividing the grams by the atomic weight of the element. Many times students will want to use 2.016 for hydrogen, thinking that it is H2. This is wrong, use 1.008 for H.
3) Divide each molar amount by the lowest value, seeking to modify the molar amounts into small, whole numbers.
Metadatics 1 4 3 – powerful audio metadata editor. Steps 2 and 3 are the technique for determining the empirical formula. Step one is required because you have all your carbon, for example, in the form of CO2 instead of a simpler problem where it tells you how much carbon is present.
Finally, a common component of this type of problem is to provide the molecular weight of the substance and ask for the molecular formula. For example, the empirical formula of benzene is CH while the molecular formula is C6H6. Several of the problems below include this question and you can go here for a discussion about calculating the molecular formula once you know the empirical formula.
Example #1: A 1.50 g sample of hydrocarbon undergoes complete combustion to produce 4.40 g of CO2 and 2.70 g of H2O. What is the empirical formula of this compound?
Solution:
1) Determine the grams of carbon in 4.40 g CO2 and the grams of hydrogen in 2.70 g H2O.
carbon: 4.40 g x (12.011 g / 44.0098 g) = 1.20083 ghydrogen: 2.70 g x (2.0158 g / 18.0152 g) = 0.3021482 g
2) Convert grams of C and H to their respective amount of moles.
carbon: 1.20083 g / 12.011 g/mol = 0.09998 molhydrogen: 0.3021482 g / 1.0079 g/mol = 0.2998 mol
3) Divide each molar amount by the lowest value, seeking to modify the above molar amounts into small, whole numbers.
![Grams Grams](https://mac-cdn.softpedia.com/screenshots/iina_14.jpg)
hydrogen: 0.2998 mol / 0.09998 mol = 2.9986 = 3
4) We have now arrived at the answer:
the empirical formula of the substance is CH3
Note: I did not check for the presence of oxygen. The problem said hydrocarbon, which are compounds with only C and H. Sometimes the problem will be silent about what type of compound it is, give only C and H data, but oxygen will also be in the compound. See comment in Example #3.
Example #2: A 0.250 g sample of hydrocarbon undergoes complete combustion to produce 0.845 g of CO2 and 0.173 g of H2O. What is the empirical formula of this compound?
Solution:
1) Determine the grams of C in 0.845 g CO2 and the grams of H in 0.173 g H2O.
carbon: 0.845 g x (12.011 g / 44.0098 g) = 0.2306 ghydrogen: 0.173 g x (2.0158 g / 18.0152 g) = 0.01935 g
2) Convert grams of C and H to their respective amount of moles.
carbon: 0.2306 g / 12.011 g / mol = 0.01920 molhydrogen: 0.01935 g / 1.0079 g/mol = 0.01921 mol
3) Divide each molar amount by the lowest value, seeking to modify the above molar amounts into small, whole numbers.
carbon: 0.01920 mol / 0.01920 mol = 1hydrogen: 0.01921 mol / 0.01920 mol = 1
4) We have now arrived at the answer:
the empirical formula of the substance is CH.
Example #3: A 0.2500 g sample of a compound known to contain carbon, hydrogen and oxygen undergoes complete combustion to produce 0.3664 g of CO2 and 0.1500 g of H2O. What is the empirical formula of this compound?
Solution:
1a) Determine the grams of carbon in 0.3664 g CO2 and the grams of hydrogen in 0.1500 g H2O.
carbon: 0.3664 g x (12.011 g / 44.0098 g) = 0.1000 ghydrogen: 0.1500 g x (2.0158 g / 18.0152 g) = 0.01678 g
1b) Determine the grams of oxygen in the sample by subtraction.
0.2500 − (0.1000 g + 0.01678) = 0.1332 g
Notice that the subtraction is the mass of the sample minus the sum of the carbon and hydrogen in the sample. Also, it is quite typical of these problems to specify that only C, H and O are involved.
A warning: sometimes the problem will give the CO2 and H2O values, but FAIL to say that C, H, and O are involved. Make sure you add the C and H values (or sometimes the C, H, and N values) and check against the mass of the sample. Any difference would be an amount of oxygen present (or it might be a mistake!! Keep double checking your work as you do each calculation.)
2) Convert grams of C, H and O to their respective amount of moles.
carbon: 0.1000 g / 12.011 g / mol = 0.008325 molhydrogen: 0.01678 g / 1.0079 g/mol = 0.01665 mol
oxygen: 0.1332 g / 15.9994 g/mol = 0.008327 mol
3) Divide each molar amount by the lowest value, seeking to modify the molar amounts into small, whole numbers.
carbon: 0.008325 mol / 0.008325 mol = 1hydrogen: 0.01665 mol / 0.008325 mol = 2
oxygen: 0.008327 mol / 0.008325 mol = 1
4) We have now arrived at the answer:
the empirical formula of the substance is CH2O
Example #4: Quinone, which is used in the dye industry and in photography, is an organic compound containing only C, H, and O. What is the empirical formula of the compound if you find that 0.105 g of the compound gives 0.257 g of CO2 and 0.0350 g of H2O when burned completely? Given a molecular weight of approximately 108 g/mol, what is its molecular formula?
Iina 1 0 6 Grams =
Comment: as a reminder, the following link goes to a discussion of how to calculate the molecular formula once you get the empirical formula.
Solution:
1) mass of each element:
carbon ⇒ 0.257 g x (12.011 / 44.0098) = 0.07014 g
hydrogen ⇒ 0.0350 g x (2.016 / 18.015) = 0.00391674 g
oxygen ⇒ 0.105 g − (0.07014 + 0.00391674) = 0.03094326 g
2) moles of each element:
carbon ⇒ 0.07014 g / 12.011 g/mol = 0.005840 mol
hydrogen ⇒ 0.00391674 g / 1.008 g/mol = 0.0038856 mol
oxygen ⇒ 0.03094326 g / 16.00 g/mol = 0.0019340 mol
3) Look for smallest whole-number ratio:
carbon ⇒ 0.005840 / 0.0019340 = 3
hydrogen ⇒ 0.0038856 / 0.0019340 = 2
oxygen ⇒ 0.0019340 / 0.0019340 = 1
4) Empirical formula:
C3H2O
5) Molecular formula:
the weight of C3H2O is 54108 / 54 = 2
C6H4O2
Example #5: A 1.000 g sample of a compound is combusted in excess oxygen and the products are 2.492 g of CO2 and 0.6495 g of H2O.
a) Determine the empirical formula of the compound.
b) Given that its molar mass is 388.46 g/mol, determine the compound's molecular formula.
Solution:
1) mass of each element:
carbon ⇒ 2.492 g x (12.011 / 44.0098) = 0.68011 ghydrogen ⇒ 0.6495 g x (2.016 / 18.015) = 0.07268343 g
oxygen ⇒ 1.000 − (0.68011 + 0.07268343) = 0.24720657 g
Notice that there was oxygen in the compound and that the problem did not tell you that.
2) moles of each element:
carbon ⇒ 0.68011 g / 12.011 g/mol = 0.0566240 mol
hydrogen ⇒ 0.07268343 g / 1.008 g/mol = 0.0721066 mol
oxygen ⇒ 0.24720657 g / 16.00 g/mol = 0.01545 mol
3) Look for smallest whole-number ratio:
carbon ⇒ 0.0566240 / 0.01545 = 3.665hydrogen ⇒ 0.0721066 / 0.01545 = 4.667
oxygen ⇒ 0.01545 / 0.01545 = 1
Do NOT round these off. You should only round off with numbers very close to a whole number. How close? Something like 2.995 goes to 3.
4) I want to change the numbers to improper fractions
carbon ⇒ 3.665 = 11/3hydrogen ⇒ 4.667 = 14/3
Iina 1 0 6 Grams Ounces
oxygen ⇒ 1 = 3/3
Sometimes, a textbook will 'magically' tell you what factor to multiply by and you will wonder why that particular factor was selected. Notice that 3.665 is three-and-two-thirds, 4.667 is four-and-two-thirds. I changed everything to fractions to try and highlight why three is used.
5) Multiply by three to get the whole-number ratio:
11 : 14 : 3empirical formula = C11H14O3
6) The weight of the empirical formula is 194:
388 / 194 = 2the molecular formula is C22H28O6
Example #6: A carbohydrate is a compound composed solely of carbon, hydrogen and oxygen. When 10.7695 g of an unknown carbohydrate (MW = 128.2080 g/mol) was subjected to combustion analysis with excess oxygen, it produced 29.5747 g CO2 and 12.1068 g H2O. What is its molecular formula?
Solution:
1) Carbon:
mass of CO2 = 29.5747 g ⇒ 0.6722 moles of CO2 ⇒ 0.6722 moles of C ⇒ 8.066 g of C
2) Hydrogen:
mass of H2O = 12.107 g ⇒ 0.67260 moles of H2O ⇒ 1.3452 moles of H ⇒ 1.3558 g of H
3) Oxygen:
mass of compound burnt = 10.770 g mass of C + H = 9.422 g
10.770 g − 9.422 g = 1.348 g of O ⇒ 0.08424 mol of O
4) Determine empirical and molecular formula:
molar ratio of C : H : O ⇒ 0.6722 : 1.3452 : 0.08424 after dividing by the smallest
molar ratio of C : H : O ⇒ 7.98 : 15.97 : 1.00 <--- the first two are close enough so as to be rounded off
empirical formula is C8H16O
'empirical formula weight' = 96 + 16 + 16 = 128 which is the molecular weight so the molecular formula is also
C8H16O
Note: I use 'EFW' for the term 'empirical formula weight.' Neither is in standard usage in the world of chemistry.
Example #7: Caffeine, a stimulant found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 1.000 mg of caffeine produces 1.813 mg CO2, 0.4639 mg H2O, and 0.2885 mg N2. Estimate the molar mass of caffeine, which lies between 150 and 200 g/mol.
Example #8: A 1.50 gram sample of cadaverine gives 3.23 g of CO2, 1.58 g of N2O5, and 1.865 g of H2O. Its molar mass is 102.2 g/mol. Determine the empirical and molecular formulas.
Example #9: Lysine is an amino acid which has the following elemental composition: C, H, O, N. In one experiment, 2.175 g of lysine was combusted to produce 3.94 g of CO2 and 1.89 g H2O. In a separate experiment, 1.873 g of lysine was burned to produce 0.436 g of NH3. The molar mass of lysine is approximately 150 g/mol. Determine the empirical and molecular formula of lysine.
Example #10: Compound A contains 5.2% by mass of nitrogen as well as C, H and O. Combustion of 0.0850 g of compound A gave 0.224 g of CO2 and 0.0372 g of H2O. Calculate the empirical formula of A.
Determine the Empirical Formula Using Combustion Analysis |
Stoichiometry
Mass-Mass Examples
This is the most common type of stoichiometric problem in high school.
There are four steps involved in solving these problems:
- Make sure you are working with a properly balanced chemical equation.
- Convert grams of the substance given in the problem to moles.
- Construct two ratios - one from the problem and one from the chemical equation and set them equal. The ratio from the problem will have an unknown, 'x.' Solve for 'x.'
- Convert moles of the substance just solved for into grams.
Comments
- Double check the equation. The ChemTeam has seen lots of students go right ahead and solve using the unbalanced equation supplied in the problem (or test question for that matter).
- DON'T use the same molar mass in steps two and four. Your teacher is aware of this and, on a multiple choice test, will provide the answer arrived at by making this mistake. You have been warned!
- Don't multiply the molar mass of a substance by the coefficient in the problem BEFORE using it in one of the steps above. For example, if the formula says 2H2O in the chemical equation, DON'T use 36.0 g/mol, use 18.0 g/mol.
- Don't round off until the very last answer. In other words, don't clear your calculator after step two and write down a value of 3 or 4 significant figures to use in the next step. Round off only once after all calculations are done.
STOP!!!
Go back to the start of this file and re-read it. Notice that I give four steps (and some advice) in how to solve the example problems just below. My advice is to keep going back to those steps as you examine the examples below.
Example #1: How many grams of hydrogen gas are needed to react completely with 54.0 g of oxygen gas, given the following unbalanced chemical reaction:
H2 + O2 ---> H2O
Solution:
1) Balance the chemical equation:
2H2 + O2 ---> 2H2O
2) Convert grams of the substance given:
54.0 g / 32.0 g/mol = 1.6875 mol of O2Note the use of 32.0 and not 16.0. The chemical substance is O2. Students have been known to sometimes forget to write the subscript of 2 on a diatomic element (H2, N2, O2, F2, Cl2, Br2, I2)
3) Construct two molar ratios and set them equal to each other:
First molar ratio is from the coefficients of the balanced chemical equation. The two substances are:H2 |
––– |
O2 |
and the numerical ratio is this:
2 |
––– |
1 |
The second ratio is found within the problem statement. The H2 is our unknown because the problem says 'how many grams of hydrogen' and the O2 mole amount is the other value. Like this:
x |
––––– |
1.6875 |
I left the mol unit off for convenience. Note also that I did not round off. I'll do that at the end.
We need to set the two ratios equal to each other and solve:
2 | x | |
––– | = | ––––– |
1 | 1.6875 |
x = 3.375 mol of H2 required
4) Convert the calculated moles from step #3 into grams:
(3.375 mol) (2.016 g/mol) = 6.80 g (to three sig figs)
Note: if you did not balance the equation, you'd wind up using an incorrect 1:1 molar ratio rather than the correct 2:1 ratio.
Example #2: How many grams of hydrogen gas are needed to produce 105.0 grams of water, given the following unbalanced chemical reaction:
H2 + O2 ---> H2O
Solution:
1) Balance the chemical equation:
2H2 + O2 ---> 2H2O
2) Convert grams of the substance given:
105.0 g / 18.015 g/mol = 5.82848 mol of H2OI rounded off some, but I made sure to keep more digits than what I will round off to at the end.
3) Construct two molar ratios and set them equal to each other:
The two substances in our ratios are these:H2 |
–––– |
H2O |
and the numerical ratio from the coefficients of the chemical equation is this:
2 |
––– |
2 |
The second ratio comes from information in the problem:
x |
––––– |
5.82848 |
Setting equal and solving:
2 | x | |
––– | = | ––––––– |
2 | 5.82848 |
x = 5.82848 mol of H2 required
Note: this can be an area of confusion. Since the ratio is a 1:1 ratio, the answer of 5.82848 mol is arrived at easily. However, many students will forget that the 5.82848 mol answer is now that of the OTHER substance, the hydrogen.
It seems that, because the number (the 5.82848) didn't change, the substance didn't change. Consequently, the student will enter the next (and last) step thinking the 5.82848 still refers to water.
4) Convert moles to grams:
(5.82848 mol) (2.016 g/mol) = 11.75 g of H2 (to four sig figs)
Example #3: How many grams of hydrogen gas are needed to produce 85.2 grams of ammonia, given the following unbalanced chemical reaction:
N2 + H2 ---> NH3
Solution:
1) Balance the chemical equation:
N2 + 3H2 ---> 2NH3
2) Convert the given grams to moles:
85.2 g / 17.0307 g/mol = 5.00273 mol
3) Construct two molar ratios and set them equal to each other:
The two substances in our ratios are these:H2 |
–––– |
NH3 |
The two ratios set equal to each other are:
3 | x | |
––– | = | ––––––– |
2 | 5.00273 |
x = 7.504095 mol of H2
4) Convert the calculated moles to grams: Sweet home 3d 6 1 32.
(7.504095 mol) (2.016 g/mol) = 15.8 g (to three sig figs)
Example #4: How many grams of chlorine can be liberated from the decomposition of 64.0 g. of AuCl3 by this reaction:
AuCl3 ---> Au + Cl2
Solution:
1) The provided equation must be balanced correctly:
2AuCl3 ---> 2Au + 3Cl2
2) Convert grams of AuCl3 to moles:
Let x = the moles of AuCl364.0 g | |
x = | –––––––––––– |
303.32 g/mol |
x = 0.210998 mol of AuCl3
Window focus 1 0 4 full crack mac os x. The ChemTeam has heard many variations of this:
'But how did you know to convert grams of AuCl3 to moles?'
![Iina 1 0 6 grams equals Iina 1 0 6 grams equals](https://mac-cdn.softpedia.com/screenshots/iina_9.jpg)
I picked AuCl3 to convert from grams to moles because a gram amount of AuCl3 was provided in the problem.
3) Use two molar ratio involving AuCl3 and Cl2:
AuCl3 |
––––– |
Cl2 |
The two molar ratios set equal to each other:
2 | 0.210998 | |
––– | = | ––––––– |
3 | x |
x = 0.316497 mol of Cl2
This is the hardest step. Constructing the proper ratio and proportion causes a great deal of confusion.
4) Convert the calculated moles to grams:
(0.316497 mol) (70.906 g/mol) = 22.4 g (to three sig figs)
One question I often get is 'Where did the value of 303.32 come from?' Answer - it's the molar mass of AuCl3. Keep this answer in mind as you wonder about where other numbers come from in a given solution.
Iina 1 0 6 Grams Equals
You might also want to consider looking at the solution to the problem and try to fit it to the list of steps given above. I know what I am suggesting is horrible and very mean, but then, I'm a teacher. What the heck do I know?
Example #5: Calculate the mass of AgCl that can be prepared from 200. g of AlCl3 and sufficient AgNO3, using this equation:
3AgNO3 + AlCl3 ---> 3AgCl + Al(NO3)3
Solution:
1) Since the chemical equation is already balanced, let us convert grams of AlCl3 to moles:
200. g | |
–––––––––––– | = 1.499914 mol of AlCl3 |
133.341 g/mol |
I picked AlCl3 because it was the substance has a gram amount associated with it in the problem.
2) Use a proportion with molar ratios involving AgCl and AlCl3:
AgCl |
––––– |
AlCl3 |
3 | x | |
––– | = | ––––––– |
1 | 1.499914 |
x = 4.499742 mol of AgCl
The 'x' in the right-hand ratio is associated with the substance we are trying to calculate an amount for (the AgCl). Look for phrases like 'Calculate the mass of . . .' or 'Determine the mass of . . . ' in the problem statement.
3) Convert moles to grams:
(4.499742 mol) (143.323 g/mol) = 645 g (to three sig figs)
By the way, what if you had used the ratio of 1 over 3, with the AlCl3 value in the numerator? Then, the other ratio would have been reversed and the answer would have been the same. The ratio and proportion would have looked like this:
1 | 1.499914 | |
––– | = | ––––––– |
3 | x |
Example #6: Given this equation:
2KI + Pb(NO3)2 ---> PbI2 + 2KNO3
calculate mass of PbI2 produced by reacting of 30.0 g KI with excess Pb(NO3)2
Solution:
1) The equation is balanced. Sometimes you're given an unbalanced equation on the test when all the classroom examples used already-balanced equations. Make sure you do these problems with a balanced chemical equation.
2) We are given 30.0 g of KI. Change it to moles:
30.0 g | |
–––––––––––– | = 0.180725 mol of KI |
165.998 g/mol |
3) Construct a ratio and proportion:
This ratio:2 |
––– |
1 |
comes from the coefficients of the balanced equation.
This ratio:
0.180725 |
–––––––– |
x |
comes from a consideration of the data in the problem.
Setting the two ratios equal to each other gives us the proportion to solve:
2 | 0.180725 | |
–– | = | –––––––– |
1 | x |
x = 0.0903625 mol <--- this is moles of PbI2
The substance associated with the 'x' is not the one for which the grams are given in the problem statement. The 'x' is associated with the substance for which a phrase like 'Determine how much . . .' is used.
Notice that a third substance (the Pb(NO3)2) is mentioned, but the word excess is used to describe it. As you learn more about stoichiometry, the excess substance will be brought into the calculations. Not yet, however. Look for it in a section called 'limiting reagent.'
4) Convert moles to grams:
(0.0903625 mol) (461.01 g/mol) = 41.6 g (to three sig figs)
Example #7: If 92.0 g of aluminum is produced, how many grams of aluminum nitrate reacted?
Al(NO3)3 + Mg ---> Mg(NO3)2 + Al
Solution:
1) An unbalanced equation was given in the problem. It needs to be balanced:
2Al(NO3)3 + 3Mg ---> 3Mg(NO3)2 + 2Al
2) Grams of aluminum is given. Convert it to moles:
92.0 g | |
–––––––––– | = 3.4099 mol of Al |
26.98 g/mol |
3) Use a ratio and proportion involving aluminum and aluminum nitrate:
Al |
–––––––– |
Al(NO3)3 |
2 | 3.4099 | |
–– | = | ––––––– |
2 | x |
x = 3.4099 mol <--- this is moles of Al(NO3)3, NOT moles of Al
Warning: there will be a real temptation in the next step to use the wrong molar mass
4) Determine grams of the unknown, the aluminum nitrate:
(3.4099 mol) (212.994 g/mol) = 726 g (to three sig figs)
Comments about the ending step of Example #7:
It is quite common in a problem like this for the student to use the molar mass of Al in this step. I think it is because they see the same value (the 3.4099 mol) in this step as in the second step. The conclusion is that it must be the same substance. And that is in error.
In the second step, we had 3.4099 mol of aluminum, but after solving the ratio and proportion, we now have 3.4099 mol of aluminum nitrate.
Be careful on the point, especially if the amount you got at the end equals the amount you had at the beginning (the 92 grams).
Example #8: How many grams of AuCl3 can be made from 100.0 grams of chlorine by this reaction:
2Au + 3Cl2 ---> 2AuCl3
Solution:
1) The equation is balanced. Yay!
2) 100.0 g of chlorine is given in the problem. Convert it to moles:
100.0 g | |
–––––––––– | = 1.41032 mol of Cl2 |
70.906 g/mol |
Notice that the element chlorine is diatomic. Students sometimes forget to write the seven diatomics with the subscripted two. The seven diatomics are: H2, N2, O2, F2, Cl2, Br2, I2
3) The ratio and proportion will involve Cl2 and AuCl3:
3 | 1.41032 mol | |
–– | = | ––––––––––– |
2 | x |
x = 0.940213 mol
Notice that the values associated with chlorine (3 and 1.41032) are in the numerator and the values associated with gold(III) chloride (2 and x) are in the denominator. If you were to flip one ratio, you'd have to flip the other.
4) Convert moles of AuCl3 to grams:
(0.940213 mol) (303.329 g/mol) = 285 g
Example #9: Aluminum foil 1.00 cm square and 0.540 mm thick react with bromine to form aluminum bromide. (a) How many grams of bromine were consumed? (b) How many grams of aluminum bromide were produced?
Solution:
1) Let us determine the mass, then moles, of Al present:
volume of Al foil ---> (1.00 cm) (1.00 cm) (0.0540 cm) = 0.0540 cm3Note the change of mm to cm.
mass of Al ---> (2.70 g/cm3) (0.0540 cm3) = 0.1458 g
Note the use of the density of aluminum.
moles of Al ---> 0.1458 g / 26.98154 g/mol = 0.0054037 mol
2) The equation for the reaction is this:
2Al + 3Br2 ---> 2AlBr3The Al to Br2 molar ratio of 2:3 will be used to answer (a). The Al to AlBr3 molar ratio of 2:2 will be used to answer (b).
3) Use the Al to Br2 molar ratio to determine moles of Br2 consumed:
2 | 0.0054037 mol | |
–––– | = | ––––––––––––– |
3 | x |
x = 0.00810555 mol (of Br2)
4) Determine grams of Br2:
(0.00810555 mol) (159.808 g/mol) = 1.30 g (to three sig figs)
5) Use the Al to AlBr3 molar ratio to determine moles of AlBr3 produced:
2 | 0.0054037 mol | |
–––– | = | ––––––––––––– |
2 | x |
x = 0.0054037 mol (of AlBr3)
6) Determine grams of AlBr3:
(0.0054037 mol) (266.694 g/mol) = 1.44 g (to three sig figs)
Example #10: How many grams of oxygen are in a sample of Ca3(PO4)2 that contains 66.0 g of calcium?
Comment: stoichiometric problems are usually of the 'I have one chemical substance, how much of another chemical substance'? variety. But, they don't have to be. Here is an example of a mass-mass stoichiometric problem based on the relationships within one chemical substance.
Solution:
1) Determine moles of calcium:
66.0 g / 40.078 g/mol = 1.6468 mol
2) Determine moles of oxygen in the sample, based on a 3:8 ratio between Ca and O:
3 | 8 | |
––––––––– | = | ––– |
1.6368 mol | x |
x = 4.3648 mol
3) Determine mass of oxygen:
(4.3648 mol) (16.00 g/mol) = 69.8 g
Bonus Example: Solid lithium hydroxide is used in space vehicles to removed exhaled carbon dioxide from the living environment by forming solid lithium carbonate and liquid water. (a) What mass of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium? (b) At STP, what is the volume of CO2 produced?
Solution:
1) Write the balanced chemical equation for the described reaction:
2LiOH + CO2 ---> Li2CO3 + H2O
2) However, there is a possible problem. The question asks for 1.00 kg of lithium, not lithium hydroxide. We need to know the molar relationship between Li and CO2. So, let's make LiOH from Li:
2Li + 2H2O ---> 2LiOH + H2
3) If I add the two reactions, I obtain this:
2Li + CO2Iina 1 0 6 Grams Equal
+ H2O ---> Li2CO3 + H2Note that two LiOH and one H2O cancel out. This third reaction gives me the Li to CO2 as 2 to 1, so I am now ready to continue on.
4) Determine moles of Li that react:
1000 g / 6.941 g/mol = 144.07 mol
5) Using the 2:1 molar ratio, I can determine the moles of CO2 consumed:
2 | 144.07 mol | |
––– | = | ––––––––– |
1 | x |
x = 72.035 mol (of CO2)
6) Convert moles to grams to get the answer for (a):
(72.035 mol) (44.009 g/mol) = 3170 g
7) To determine the volume at STP, we can use either PV = nRT or molar volume:
PV = nRT(1.00 atm) (V) = (72.035 mol) (0.08206 L atm / mol K) (273.15 K)V = 1614.6 L (to three sig figs, this would be 1610 L)
molar volume
(22.414 L/mol) (72.035 mol) = 1614.6 L (1610 L to three sig figs)