CHM 1045
Dr. Michael Blaber

Name_________________________

SS #__________________________

1. How many grams are there in 0.5 moles of Barium Chloride? (1/2 point)

The formula for Barium Chloride is BaCl₂

therefore the formula weight is 137.3 + (2*35.5) = 208.3 g/mole

so:

0.5 moles * (208.3 g/mole) = 104.2 g

2. How many moles are there in half an ounce of silver? (assume there are 28.8 grams per ounce) (1/2 point)

The molecular weight of silver is 107.9 g/mole

0.5 ounces * (28.8 g/ounce) = 14.4 g

14.4 g * (1 mole/107.9 g) = 0.133 moles

3. For the compound sodium hydroxide (NaOH) what is the percentage by mass for each element Na, O and H? (2 points)

The molecular weight of Na is 23.0 g/mole

The molecular weight of O is 16.0 g/mole

The molecular weight of H is 1.01 g/mole

Therefore, the mass of 1.0 mole of NaOH would be (23.0 + 16.0 + 1.01) = 40.01 g

The mass % of Na would be (23.0/40.01) x 100 = 57.5%

The mass % of O would be (16.0/40.01) x 100 = 40.0%

The mass % of H would be (1.01/40.01) x 100 = 2.5%

4. A certain hydrocarbon compound (containing only C and H) is 83.6% C and 16.4% H by mass. The experimentally determined molecular mass is 86.2 amu. What is the empirical and chemical formula for this hydrocarbon? (2 points)

Choosing 100g as a convenient mass, we would have 83.6 g of C and 16.4 g of H

83.6 g C * (1 mole/12.01g) = 6.96 moles C

16.4 g H * (1 mole/1.01g) = 16.24 moles H

Calculating the relative stoichiometry by dividing by the smallest:

(6.96/6.96) = 1.00 moles C

(16.24/6.96) = 2.33 moles H

The value for H looks like 2 1/3, so multiply both molar amounts by 3 to give an empirical formula of C₃H₇

This would have an atomic mass of (3*12.01)+(7*1.01) = 43.1 amu

The actual molecular mass is 86.2 amu, or (83.6/43.1) = 2.0 times that of the empirical formula. Therefore, the chemical formula must be: C₆H₁₄

5. Hexane has the chemical formula C₆H₁₄ . If 5.8 grams of hexane is combusted to completion, producing only CO₂ and H₂O, how many grams each of CO₂ and H₂O are produced? (2 points)

From the chemical formula hexane has a molecular mass of 86.2 g/mole

8. g * (1 mole/86.2g) = 0.0673 moles hexane

The balanced equation for the combustion would be:

2C₆H₁₄ + 19O₂ -> 12CO₂ + 14H₂O

The stoichiometric relationships therefore are:

0.0673 moles hexane * (12 moles CO₂ / 2 moles hexane) = 0.404 moles CO₂

0.0673 moles hexane * (14 moles H₂O / 2 moles hexane) = 0.471 moles H₂O

The masses of CO₂ and H₂O produced are therefore:

0.404 moles CO₂ * (44.0 g/mole) = 17.8 g CO₂

0.471 moles H₂O * (18.0 g/mole) = 8.48 g H₂O

6. In the combustion of hexane from the above example, how many grams of CO₂ would be produced if 11.6 g of hexane were combusted with 30 g of O₂ ? Which is the limiting reagent? (2 points)

11.6 g hexane * (1 mole/86.2 g) = 0.135 moles hexane

30g O₂ * (1 mole/32g) = 0.938 moles O₂

if O₂ is unlimited:

0.135 moles hexane * (12 moles CO₂ / 2 moles hexane) = 0.81 moles CO₂

if hexane is unlimited:

0.938 moles O₂ * (12 moles CO₂ / 19 moles O₂) – 0.592 moles CO₂

Therefore, the limiting reagent is the oxygen

With the given amounts of hexane and oxygen we can make:

0.592 moles CO₂ * (44.01 g/mole) = 26.1 g CO₂

7. What is the molarity of a solution of NaCl made by dissolving 12.3 g of NaCl in 130 mls of H₂O? (1/2 point)

12.3 g NaCl * (1 mole/58.5 grams) = 0.21 moles NaCl

0.21 moles / 0.13 liters = 1.62 moles / liter (i.e. 1.62 M)

8. You have a stock solution of 5 M NaCl and an unlimited amount of pure water. How would use these two to make a 2.8 liter solution of 0.15 molar NaCl? (1/2 point)