2. Unit 4 Practice Test Chapter 7 Concepts (Mole Concept) (Percent Composition) (Empirical Formulas)

3. 1. Convert 53.75 g Cu(NO 3 ) 2 to moles of copper (II) nitrate. 53.75 g Cu(NO 3 ) 2 = mol Cu(NO 3 ) 2 g Cu(NO 3 ) 2 1 mol Cu(NO 3 ) 2 X 1 mole = molar mass (Use the Periodic Table)

4. 53.75 g Cu(NO 3 ) 2 = mol Cu(NO 3 ) 2 g Cu(NO 3 ) 2 1 mol Cu(NO 3 ) 2 X + 96.00 g =187.57 g Cu(NO 3 ) 2 63.55 g x 1 14.01 g x 2 63.55 g+ 28.02 g 16.00 g x 6 187.57 0.2866 29 Cu 63.55 8 O 16.00 7 N 14.01 On calculator: 53.75 ÷ 187.57 = 0.2865596844

5. 2. How many molecules of carbon tetrachloride are in 2.58 moles of CCl 4 ? 2.58 mol CCl 4 = molecules CCl 4 1 mol CCl 4 molecules CCl 4 X 1 mole = 6.02 x 10 23 molecules 6.02 x 10 23 1.55 x 10 24 On calculator: 2.58 x 6.02 E 23 = 1.55316 E 10 24

6. 3. What volume (in Liters) of ammonia gas is in 2.57 moles of NH 3 at STP? 2.57 mol NH 3 = L NH 3 1 mol NH 3 L NH 3 X 1 mole = 22.4 L for gas @STP 22.4 57.6 On calculator: 2.55 x 22.4 = 57.568

7. %CX 100 = 80.0 g 44.0 g = 55 % C 4.A sample of a compound containing carbon and hydrogen has a mass of 80.0 g. Experimental procedures show that 44.0 g is carbon, and the remaining 36.0 g is hydrogen. What is the percentage of Carbon in this sample?

8. 134.45 g CuCl 2 CuCl 2 63.55 g35.45 g x 2 63.55 g+ 70.90 g = x 1 %ClX 100 = 134.45 g 70.90 g = 52.73 % Cl 5. What is the percentage of chlorine in CuCl 2 ? 29 Cu 63.55 17 Cl 35.45

9. 6.0 g C= 12.01 g C 1 mol C x 0.50 mol C 16.0 g O= 16.00 g O 1 mol O x 0.50 mol = 1 = 2 CO 2 6 C 12.01 6.Find the empirical formula of a gas that contains 6.0 g carbon and 16.0 g of oxygen. Name the gas. 8 O 16.00 Carbon dioxide Name: ___________________ Empirical Formula: ______________