Empirical Formulae and Percent Composition
1. Calculate the mass percentage of each element in the following compounds:
a. Rh(CN)4
b. Co(NO2)2
2. Calculate the mass proportions, expressed as percentage composition, for each of the
elements in the compounds below:
a. Ca2P2O7
b. Be2SiO4
3. Analysis of a 15.0 g sample of brass reveals it contains 13.0 g of copper.
a) What is the mass percentage of copper in the sample?
b) How many grams of brass would yield 95.0 g of pure copper?
c) The only other component of this alloy is zinc. How many grams of zinc would be
needed to prepare 600. g of this type of brass?
4. Calcium, nitrogen and oxygen were found to be present in a compound in the
following mass percentages: 24.4% Ca, 17.1% N and 58.5% oxygen. Determine the
empirical formula of the compound.
5. A compound of carbon and fluorine is found to contain 17.4% carbon and 82.6%
fluorine.
a) How many moles of carbon atoms are there in a 100. g sample?
b) How many moles of fluorine atoms are there in a 200. g sample?
c) What is the simplest formula for this compound?
6. Calculate the empirical formula for each compound:
a) 82.6% C, 17.4% H
b) 7.20% P, 92.8% Br
c) 38.4% Mn, 16.8% C, 44.8% O
d) 43.7% Zr, 10.3% B, 46.0% O
7. The composition of 2,4,6-trinitrotoluene (TNT) is 37.0% C, 2.22% H, 18.5% N, and
42.3% O. The molar mass of TNT is 227 g/mole. What is its molecular formula?
8. A 13.4 g sample of a compound contains 5.36 g of carbon, 0.902 g of hydrogen and
7.14 g of oxygen. Determine the empirical formula of the compound.
Empirical Formulas from chemical reactions
1) A 36 g sample of iron oxide is composed of 78% iron. What is the empirical formula
of this iron oxide?
2) 15.53 g of rust (FexOy) is decomposed to produce 10.87 g iron. What is the empirical
formula of rust?
3) 72.06 g of carbon react with hydrogen gas to produce 88.06 g of propane. What is
the empirical formula of propane?
4) 58.5 g C reacts with hydrogen gas producing 73.0 g ethane. What is the formula of
ethane?
5) 20.16 g of Vanadium Oxide decomposes to give 8.88 g of Oxygen. What is the
formula of this Vanadium Oxide?
Answers to Empirical Formulae and Percent Composition
1.
a) Molar mass of Rh(CN)4 = 206.9 g/mole
% MassRh = ((1 x 102.9 g/mole) / 206.9 g/mole) x 100% = 49.7% Rh
% MassC = (4 x 12.0 g/mole) / 206.9 g/mole x 100% = 23.2% C
% MassN = (4 x 14.0 g/mole) / 206.9 g/mole x 100% = 27.1% N
Check! Σ % Mass = 100% ???49.7% + 23.2% + 27.1% = 100.0% Good!
b) Co(NO2)2: 39.0% Co, 18.6% N, 42.4% O
2.
a) Ca2P2O7:
31.6% Ca, 24.4% P, 44.1% O
b) Be2SiO4: 16.4% Be, 25.5% Si, 58.1% O
3.
a) % MassCu = (massCu / massBrass) x 100 % = (13.0 g / 15.0 g) x
100%=% MassCu = 86.7%
b) ? gBrass = massCu / % massCu = 95.0 g / 86.7% = 110. gBrass
c) MassZn = % massZn x massBrass = (100.0% - 86.7%) (600. g) = 79.8 gZn
4.
In 100.0 g of compound, we have 24.4 g Ca, 17.1 g N, and 58.5 g O.
? moles Ca = (1 mole / 40.0 g) (24.4 g) = 0.610 moles Ca
? moles N = (1 mole / 14.0 g) (17.1 g) = 1.22 moles N
? moles O = (1 mole / 16.0 g) (58.5 g) = 3.66 moles O
Mole ratio for Ca = 0.610 moles / 0.610 moles = 1.00
Mole ratio for N = 1.22 moles / 0.610 moles = 2.00
Mole ratio for O = 3.66 moles / 0.610 moles = 6.00
The formula is CaN2O6, which we recognise as Ca(NO3)2
5. a) 1.45 moles C b) 8.69 moles F c) CF3
6. a) C2H5
b) PBr5
c) MnC2O4
d) Zr(BO3)2
7.
Empirical formula: C7H5N3O6
Molecular formula: C7H5N3O6
8. Empirical formula: CH2O
Answers to Empirical Formulas from chemical reactions
1)
2)
3)
4)
5)
FeO
Fe2O3
C3H8
C2H6
V2 O 5