Chemistry 10.3

“Percent Composition and Chemical Formulas”

I. Percent Composition

A. Def – the percent by mass of each element in a compound.

1. Must = 100%

2. Formula…

Grams of element X

Grams of compoundX 100%% mass of element X =

3. Ex: An 8.20 g piece of magnesium combines with 5.40 g of oxygen to form a compound. What is the % composition of this compound?

-mass of compound =

% Mg = 8.20 g

13.60 g

= 60.3%

13.60 g

X 100%

% O = 5.40 g

13.60 g

= 39.7%

*Does this make sense?

39.7% + 60.3% = 100%

X 100%

Yes

4. Ex: 9.03 g of Mg combine completely with 3.48 g of N to form a compound. What is the % composition of this compound?

5. 29.0 g Ag completely react with 4.30 g of S to form a compound. What is the % composition of this compound?

72.2 % = Mg27.8 % = N

87.1 % = Ag12.9 % = S

B. Percent Composition from the Chemical Formula

1. Formula

%Mass = grams of element in 1 mol of compound

molar mass of compound

2. Must = 100%

3. Add molar mass + # of moles

4. Ex: Calculate the % composition of propane (C3H8)

X 100%

molar mass =

% C = 36.0

44.0

% H = 8.0

44.0

44.1 g /mol

X 100% % C = 81.8%

X 100% % H = 18 %

5. Calculate the % composition of the following compounds:

a. C2H6 =

b. NaHSO4 =

c. NH4Cl =

d. CO(NH2)2 =

%C = 79.9 %H = 20.1

%Na = 19.1 %H = .8 %S = 26.7 %O = 53.3

%N = 26.2 %H = 7.6 %Cl = 66.3

%C = 20.0 %O = 26.6 %N = 46.6 %H = 6.7

II. Empirical FormalasA. Def – gives the lowest whole # ratio of atoms in a

compound.1. can be used for atoms or moles2. Ex: HO H2O2 (hydrogen peroxide)

3. Empirical formula can be the same as molecular formula.

-Ex: CO2 (1:2 ratio)4. What are the empirical formulas for the following?

a. C6H12O6

b. C4H8

c. N4O10

CH2O

CH2

N2O5

5. Ethyne (C2H2) also called acetylene, is a gas used in welder’s torches. Styrene (C8H8) is used in making polystyrene.

-What is similar about these 2 compounds?

both have the same empirical formula (CH)

B. Calculating Empirical Formulas1. Ex: What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen?

a. Step 1 – determine the # of moles (assume %’s are in grams)

b. Step 2 – divide by smallest mole #c. Step 3 – Check for decimals (2.5, 3.75, 1.25)

2. Solve (step 1)25.9 g N x 1 mol N

74.1 g O x 1 mol O

14.0 g N= 1.85 mol N

16.0 g O= 4.63 mol O

(Step 2)

1.85 mol N

1.85

4.63 mol O

1.85

= 1 mol N

= 2.5 mol O

(Step 3)

*What can we multiply by to get a whole # ratio between N and O?

= 2

1 mol N x 2 = N2

2.5 mol O x 2 = O5

Empirical Formula = N2O5

Check answer by finding % composition for N and O.

Calculate the empirical formulas for the following…

1. 94.1% O, 5.9% H

2. 79.8% C, 20.2% H

3. 67.6% Hg, 10.8% S, 21.6 O

4. 27.59% C, 1.15% H, 16.09% N, 55.17% O

OH

CH3

HgSO4

C2HNO3

III. Calculating Molecular Formulas (includes ionic compounds)

A. Determine empirical formula

B. Determine molar mass of compound

C. Divide… Molar Mass

Emp. Formula Mass

D. Take each element in the empirical formula times “step C”

E. Write out new formula

Examples:

1. Calculate the molecular formula of the compound whose molar mass is 60.0 g and empirical formula is CH4N.