1. Ch. 7 Notes -- Chemical Quantities
Review Practice Problems:
(1) How many atoms of hydrogen are there in each compound?
a) Ca(OH)2 ___ b) C3H8O___ c) (NH4)2HPO4 ___ d) HC2H3O2 ___
(2) Calculate the formula mass of each compound. (Add up all the
atomic masses for each atom from the Periodic Table.)
a) CaCO3 b) (NH4)2SO4
c) C3H6O d) Br2
e) H3PO4 f) N2O5 2 8 9 4
2 N’s = 2 x 14.0 = 28.0
8 H’s = 8 x 1.0 = 8.0
S = 32.1
4 O’s = 4 x 16.0 = 64.0
Ca = 40.1
C = 12.0
3 O’s =3 x 16.0 = 48.0
Add them up!
132.1 g/mol
Add them up!
100.1 g/mol
159.8 g/mol
C = 3 x 12.0 = 36.0
H = 6 x 1.0 = 6.0
O =16.0
2 Br’s = 2 x 79.9 =
Add them up!
58.0 g/mol
2 N’s = 2 x 14.0 = 28.0
5 O’s = 5 x 16.0 = 80.0
3 H’s = 3 x 1.0 = 3.0
P = 31.0
4 O’s = 4 x 16.0 =64.0
Add them up!
98.0 g/mol
Add them up!
108.0 g/mol

2. 3) Convert 835 grams of SO3 to moles.
4) How many molecules of CH4 are there in 18 moles?
5) How many grams of helium are there in 5.6 x 1023 atoms of helium?
6) How many molecules are there in 3.7 grams of H2O?
1 mole SO3
835 g SO3
10.4 moles of SO3 x =
80.1 g SO3
6.02 x 1023 molecules CH4
18 moles CH4
108 x 1023 molecules CH4 x =
1 mole CH4
or 1.08 x 1025 molecules CH4
4.0 grams He
5.6 x 1023 atoms He x
3.72 grams He =
6.02 x 1023 atoms He
6.02 x 1023 molecules H2O
3.7 grams H2O
1.24 x 1023 molecules H2O x =
18.0 grams H2O

3. Calculating Percent Composition by Mass
Step 1: Find the formula mass of the compound by adding the individual masses of the elements together.
Step 2: Divide each of the individual masses of the elements by the formula mass of the compound.
Step 3: Convert the decimal to a % by multiplying by 100.
Practice Problems:
(1) Find the % composition of the elements in each compound.
a) Na3PO b) SnCl4
= = 42.1%
3 Na’s = 3 x 23.0 = 69.0
P = 31.0
4 O’s = 4 x 16.0 =
÷ 164
Sn = 118.7
4 Cl’s = 4 x 35.5 =
÷ 260.7
= 45.5%
= = 18.9%
÷ 164 +
÷ 260.7
= 54.5%
260.7
= = 39.0% +
÷ 164
164

4. Elements in the Universe: % Composition by Mass

5. Earth’s Crust: % Composition by Mass

6. Entire Earth (Including Atmosphere): % Composition by Mass

7. Human Body: % Composition by Mass

8. Determining the Empirical Formula for a Compound
The empirical formula for a compound is the simplest __________ number __________ of the atoms in the compound.
Examples: H2O is the empirical formula for water.
_______ is the empirical formula for glucose, C6H12O6.
Step 1: Divide the % composition data by the atomic mass of the element. This will give you a ratio of the # of atoms in the formula.
Step 2: Divide each of these answers by the smallest ratio.
Step 3: If there is still a decimal, multiply each answer by the denominator of the “freak”, (i.e. -- multiply all the ratios by the denominator of the ratio that is still a decimal.)
[1/2= /3≈ /3≈ /4= 0.75]
whole
ratio
C1H2O1
Helpful Rhyme: % to mass, mass to mole, divide by small, times ’til whole.

9. Practice Problems:
1) An unknown compound is composed of 81.8% carbon and 18.2% hydrogen. Determine the empirical formula for the compound.
2) An unknown compound is composed of 42.9% carbon and the rest of the compound is oxygen. Determine the empirical formula for the compound.
C = 81.8% = 81.8 g
(mass to moles)
81.8 g C ÷ 12.0 = 6.82 moles
H = 18.2% = 18.2 g
18.2 g H ÷ 1.0 = 18.2 moles
(÷ by small)
(x ‘til whole)
C6.82H18.2
C1H2.67
C3H8
6.82
6.82
x 3 x 3
C = 42.9% = 42.9 g
(mass to moles)
42.9 g C ÷ 12.0 = moles
O = 57.1% = 57.1 g
57.1 g O ÷ 16.0 = moles
(÷ by small)
C3.575O3.569
C1.0O1.0 CO
3.569
3.569

10. 2) An unknown compound is composed of 42
2) An unknown compound is composed of 42.9% carbon and the rest of the compound is oxygen. Determine the empirical formula for the compound.
C = 42.9% = 42.9 g
(mass to moles)
42.9 g C ÷ 12.0 = moles C
O = 57.1% = 57.1 g
57.1 g O ÷ 16.0 = moles O
(÷ by small)
C3.575O3.569
C1.0O1.0 CO
3.569
3.569
To check your answer, you can simply find the % composition by mass of your formula…
CO = 28 g/mole
% C = 12.0/28 = 42.9%
%O = 16.0/28 = 57.1%
That matches up with the original problem!!

11. Determining the Molecular Formula for a Compound
The molecular formula for a compound is either the same as the empirical formula ratio or it is a “_________ _________ of this ratio. It represents the true # of atoms in the molecule.
Examples: 1) H2O is the empirical & molecular formula for water ) CH2O is the empirical formula for sugar, ethanoic acid, and methanol. The molecular formula for glucose is C6H12O6, (___times the empirical ratio!)
Step 1: Determine the empirical formula for the compound. (See the previous steps in the notes.)
Step 2: Calculate the empirical formula mass of the compound.
Step 3: Determine the “whole # multiple” by dividing the molecular formula mass (given in the problem) by the empirical formula mass. Multiply each of the empirical ratios by this whole number.
whole # multiple 6

12. Practice Problems:
1) An unknown compound is composed of 58.8% carbon and 9.8% hydrogen and 31.4% oxygen. The molecular formula mass is 204 g. Determine the molecular formula for the compound.
C = 58.8% = 58.8 g
(mass to moles)
58.8 g C ÷ 12.0 = 4.9 moles C
H = 9.8% = 9.8 g
9.8 g H ÷ 1.0 = 9.8 moles H
O = 31.4% = 31.4 g
31.4 g O ÷ 16.0 = 1.96 moles O
(÷ by small)
(x ‘til whole)
C4.9H9.8O1.96
C2.5H5O1
C5H10O2
x 2 x 2 x 2
1.96
1.96
1.96
(Compare)
C10H20O4
[C5H10O2 ]x 2 =
Our formula mass = 102 g
(204 ÷ 102 = 2)

13. C3.3H6.6O3.3 C1H2O1 3.3 3.3 3.3 C4H8O4 Practice Problems:
2) An unknown compound is composed of 40% carbon, 6.6% hydrogen, and 53.4% oxygen. Determine the molecular formula for the compound if the mass of one mole of the compound is 120 g.
C = 40% = 40 g
(mass to moles)
40 g C ÷ 12.0 = 3.3 moles C
H = 6.6% = 6.6 g
6.6 g H ÷ 1.0 = 6.6 moles H
O = 53.4% = 53.4 g
53.4 g O ÷ 16.0 = 3.3 moles O
(÷ by small)
C3.3H6.6O3.3
C1H2O1
3.3
3.3
3.3
(Compare)
C4H8O4
[C1H2O1 ]x 4 =
Our formula mass = 30 g
(120 ÷ 30 = 4)