Notes #18 Section Assessment 10.3

PERCENT COMPOSITION;MOLECULAR AND EMPIRICAL

FORMULAS

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%.

PERCENT COMPOSITION FROM MASS DATA

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?

o We want to know the % composition of oxygen AND magnesium.

o 1) 13.60g – 5.40g Oxygen = 8.20g magnesiumo 2) % Mg = 8.20g Mg / 13.60g Compound x 100% =

60.3%o 3) % O = 5.40g O / 13.60g Compound x100% = 39.7%

o The compound contains 60.3% Mg and 39.7% O.

EXAMPLE 1

USE THE MOLAR MASSES

PERCENT

Calculate the percent composition of C3H8.

o 1) 1 mole of carbon = 12.01 x 3 = 36.0g total for carbon

o 2) 1 mole of hydrogen = 1.01 x 8 = 8.0g total for hydrogen

o 3) Molar Mass of Compound = 3(12.01) + 8(1.01) = 44.0 g/mol

o 4) % Comp of Carbon = 36.0g / 44.0g x 100% = 81.8% carbon

o 5) % Comp of Hydrogen = 8.0g / 44.0g x 100% = 18% hydrogen

o The compound contains 81.8% carbon and 18% hydrogen.

EXAMPLE 2

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.

EMPIRICAL FORMULAS

Are either the same as the empirical formula, or it is a simple whole-number multiple of the empirical

formula.

Methanal CH2O This is the molecular formula. What is the empirical formula? CH2O. It is already

simplified.

Glucose C6H12O6 is the molecular formula. What is the empirical formula? CH2O. It can be simplified.

MOLECULAR FORMULAS

MORE EXAMPLES

Find the empirical formulas for the following molecular compounds:

o C4H8

o CH2

o C2H6O2

o X39Y13

o WO2

o C2H4NO

o C5H10O5

PRACTICE

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 compoundo 2) 74.1% O = 74.1g O per 100g of compoundo 3) Convert to moleso 4) 25.9g N x 1 mole/14.01g = 1.85 mol No 5) 74.1g O x 1 mole/16g = 4.63 mol Oo 6) 1.85 mol N/ 1.85 = 1 mol N ; 4.63 mol O/ 1.85 = 2.50 mol Oo 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: N2O5

CALCULATE EMPIRICAL FORMULAS

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

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 = 2o 4) Multiply each subscript of empirical formula by 2.o 5) C2H8N2

CALCULATE MOLECULAR FORMULAS

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.

SECTION ASSESSMENTS