Notes #18

PERCENT COMPOSITION;
MOLECULAR AND EMPIRICAL
FORMULAS
Notes #18 Section Assessment 10.3
PERCENT COMPOSITION FROM
MASS DATA
The percent by mass of an element in a compound is the
number of grams of the element divided by the mass in
grams of the compound, multiplied by 100%.
EXAMPLE 1
A 13.60g sample of a compound that contains magnesium and oxygen is
decomposed. 5.40g of oxygen is obtained. What is the percent
composition of this compound?
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We want to know the % composition of oxygen AND magnesium.
1) 13.60g – 5.40g Oxygen = 8.20g magnesium
2) % Mg = 8.20g Mg / 13.60g Compound x 100% = 60.3%
3) % O = 5.40g O / 13.60g Compound x100% = 39.7%
o The compound contains 60.3% Mg and 39.7% O.
PERCENT
USE THE MOLAR MASSES
EXAMPLE 2
Calculate the percent composition of C 3H8.
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1) 1 mole of carbon = 12.01 x 3 = 36.0g total for carbon
2) 1 mole of hydrogen = 1.01 x 8 = 8.0g total for hydrogen
3) Molar Mass of Compound = 3(12.01) + 8(1.01) = 44.0 g/mol
4) % Comp of Carbon = 36.0g / 44.0g x 100% = 81.8% carbon
5) % Comp of Hydrogen = 8.0g / 44.0g x 100% = 18% hydrogen
o The compound contains 81.8% carbon and 18% hydrogen.
EMPIRICAL FORMULAS
The lowest whole number ratio of atoms of the
elements in a compound.
Examples:
Ethyne C2H2 What is the empirical formula? CH
Styrene C8H8 What is the empirical formula? CH
They are the same! The empirical formula for a
compound does not give us much information.
MOLECULAR FORMULAS
Are either the same as the empirical formula, or it is a simple wholenumber multiple of the empirical formula.
Methanal CH2O This is the molecular formula. What is the
empirical formula? CH 2O. It is already simplified.
Glucose C6H12O6 is the molecular formula. What is the empirical
formula? CH2O. It can be simplified.
MORE EXAMPLES
PRACTICE
Find the empirical formulas for the following molecular compounds:
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C 4H 8
CH2
C 2H 6O2
X39Y13
WO2
C2H4NO
C5H10O5
CALCULATE EMPIRICAL
FORMULAS
A compound is found to contain 25.9% nitrogen and 74.1% oxygen. What is
the empirical formula for the compound?
*Percent means part per 100. Therefore,
o 1) 25.9% N = 25.9g N per 100g of compound
o 2) 74.1% O = 74.1g O per 100g of compound
o 3) Convert to moles
o 4) 25.9g N x 1 mole/14.01g = 1.85 mol N
o 5) 74.1g O x 1 mole/16g = 4.63 mol O
o 6) 1.85 mol N/ 1.85 = 1 mol N ; 4.63 mol O/ 1.85 = 2.50 mol O
o 7) Result N1O2.5. We need whole numbers
o 8) Multiply each ratio to smallest whole number (2).
o 9) 1 mol N x 2 = 2 mol N ; 2.5 mol O x 2 = 5 mol O
Answer:
N 2O 5
CALCULATE MOLECULAR
FORMULAS
Calculate the molecular formula of a compound whose molar mass is 60.0
g/mol and empirical formula is CH 4N.
o 1) Calculate the empirical formula mass: 12.01 + 4(1.01) + 14.01=30.0g
o 2) Divide the molar mass of compound by the empirical formula mass.
o 3) 60.0g / 30.0g = 2
o 4) Multiply each subscript of empirical formula by 2.
o 5) C2H8N2
SECTION ASSESSMENTS
You can now complete the three assigned section assessments!
Allow time in class to complete section 10.3 assessment. Page 312.
Due next Monday/Tuesday.
Research Paper is due Friday at 9 pm.