7.3 Percent composition and chemical formulas

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7.3 Percent composition and chemical formulas
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Percent composition

• The relative amount of mass of each element in a
  compound, expressed in

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Calculating percent composition

• composition of an element
• mass of element/mass of compound X 100
• The sum of all the compositions of all the
  elements must equal 100

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Percent composition

• We dont know the mass of the element , but we
  can find the percent composition using the
  formula and the molar mass of a substance
• percent composition

grams of element in 1 mole molar mass of compound
100
5
8.20 g Mg combines completely with 5.40 g O. What
is the percent composition of this compound?
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8.20 g Mg combines completely with 5.40 g O. What
is the percent composition of this compound?

• Add the masses of the elements
• 8.20 g 5.40 g 13.60 g
• Divide the mass of each element by the total mass
  and multiply by 100 (make it a percent)
• 8.20 g Mg / 13.60 g 60.3 Mg
• That makes the composition of O 39.7

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222.6 g N combines completely with 77.4 g O. What
is the percent composition of each element?
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222.6 g N combines completely with 77.4 g O. What
is the percent composition of each element?

• Total weight of compound is 300 g
• 222.6 g/300 g X 100 74.2 N
• 77.4 g/300 g X 100 25.8 O
• 74.2 25.8 100

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Calculate the percent composition of HCN.
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Calculate the percent composition of HCN.

• Molar mass H 1 g
• Molar mass C 12 g
• Molar mass N 14 g
• Total molar mass 27 g
• 1 g H/27 g X 100 3.7 H
• 12 g C/27 g X 100 44.4 C
• 14 g N/27 g X 100 51.9 N

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Using the composition of H in HCN (3.7),
calculate the amount of H in 378 g HCN.
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Using the composition of H in HCN (3.7),
calculate the amount of H in 378 g HCN.

• 378 g HCN X 3.7 g H/100 g HCN

378 g HCN
3.7 g H 100 g HCN
14 g H
Conversion factor There are 3.7 g of H in each
100 g HCN
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Empirical formulas

• Percent composition can be used to calculate the
  empirical formula for a compound
• Empirical formulas are the lowest whole number
  ratios of the atoms in a compound
• Empirical formulas may or may not be the actual
  formula when we are dealing with molecules
• For example, hydrogen peroxide, H2O2 has an
  empirical formula of HO, but doesnt occur in
  nature that way

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Other cases where empirical formulas dont tell
us the composition of a molecule

• Empirical formula CH
• Molecular formula C2H2 (ethyne or acetylene)
• Molecular formula C6H6 (benzene)
• Empirical formula CH2O
• Molecular formula C2H4O2 (ethanoic acid)
• Molecular formula C6H12O6 (glucose)

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What is the empirical formula of a compound which
is 25.9 N and 74.1 O?
25.9 g N
1 mol N 14 g N
1.85 mol N
74.1 g O
1 mol O 16 g O
4.63 mol O
The ratio of moles of N to moles of O is
1.85/4.63, or 1/2.5. In whole numbers, this is a
ratio of 2/5. Therefore, the empirical formula
for this compound is N2O5.
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Calculate the empirical formula of a compound
which is 43.64 P and 56.36 O.
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Calculate the empirical formula of a compound
which is 43.64 P and 56.36 O.
43.64 g P
1 mol P 30.97 g P
1.409 mol P
56.36 g O
1 mol O 16 g O
3.523 mol O
The ratio of moles of P to moles of O is
1.409/3.523, or 1/2.5. In whole numbers, this is
a ratio of 2/5. Therefore, the empirical formula
for this compound is P2O5.
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A compound is 2.477 g Mn and 1.323 g O. What is
the empirical formula?
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A compound is 2.477 g Mn and 1.323 g O. What is
the empirical formula?
2.477 g Mn
1 mol Mn 59.94 g Mn
.0413 mol Mn
1.323 g O
1 mol O 16 g O
.0827 mol O
The ratio of moles of Mn to moles of O is
.0413/.0827, or 1/2. Therefore, the empirical
formula for this compound is MnO2.
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Methyl butanoate has a percent composition
of58.8 C 9.8 H and 31.4 O. Its molar
mass is 102 grams. Want is its molecular formula?
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Methyl butanoate has a percent composition
of58.8 C 9.8 H and 31.4 O. Its molar
mass is 102 grams. Want is its molecular formula?

• .588 X 102 g 59.98 g C
• .098 X 102 g 9.99 g H
• .314 X 102 g 32.03 g O
• These numbers look awfully suspicious!
• The first is 5 moles C, the second is 10 moles
  H, and the third is 2 moles O. So the molecular
  formula is C5H10O2