1. Notes #18 Section Assessment 10.3

2. 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

3. 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 magnesium o 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

4. USE THE MOLAR MASSES PERCENT

5. Calculate the percent composition of C 3 H 8. 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

6. The lowest whole number ratio of atoms of the elements in a compound. Examples: Ethyne C 2 H 2 What is the empirical formula? CH Styrene C 8 H 8 What is the empirical formula? CH They are the same! The empirical formula for a compound does not give us much information. EMPIRICAL FORMULAS

7. Are either the same as the empirical formula, or it is a simple whole- number multiple of the empirical formula. Methanal CH 2 O This is the molecular formula. What is the empirical formula? CH 2 O. It is already simplified. Glucose C 6 H 12 O 6 is the molecular formula. What is the empirical formula? CH 2 O. It can be simplified. MOLECULAR FORMULAS

8. MORE EXAMPLES

9. Find the empirical formulas for the following molecular compounds: o C 4 H 8 o CH 2 o C 2 H 6 O 2 o X 39 Y 13 o WO 2 o C 2 H 4 NO o C 5 H 10 O 5 PRACTICE

10. 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 N 1 O 2.5. 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 2 O 5 CALCULATE EMPIRICAL FORMULAS

11. Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH 4 N. 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) C 2 H 8 N 2 CALCULATE MOLECULAR FORMULAS

12. 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