CHM 1045
Fall 1999
Dr. Michael Blaber

Name_________________________SS #__________________________

(Due in Recitation Tues Sep 21)

1. Write the balanced chemical equation for the combustion of pentanol. (C₅H₁₂O). (1/2 point)

2C₅H₁₂O + 15O₂ -> 10CO₂ + 12H₂O

C 10 C 10

H 24 H 24

O 32 O 32

2. Potassium reacts with Chlorine to produce the ionic solid potassium chloride (KCl). Which elements would you predict would react with Bromine in a similar type of reaction? (1/2 point)

Group 1A, alkali metals, or Li, Na, Rb, Cs, and Fr

3. What is the formula weight of Sucrose (C₁₂H₂₂O₁₁)? What is the percentage by mass of C, H and O in Sucrose? (1 point)

Formula weight: (12*12.0)+(22*1.01)+(11*16.0) = 144+22.2+176.0 = 342amu

%C = 144/342 = .421=42.1%

%H = 22.2/342 = .0650=6.50%

%O = 176/342 = 0.515=51.5%

4. How many grams are there in 3.80 moles of Lithium Oxide? (1/2 point) How many moles are there in 100g of Aluminum Sulfide? (1/2 point)

The formula for Lithium Oxide would be Li₂O

Therefore the formula weight is (2*6.94) + 16.0 = 29.9 g/mole

so:

3.80 moles * (29.9 g/mole) = 114 g

The formula for Aluminum Sulfide would be Al₂S₃

Therefore the formula weight is (2*27.0) + (3*32.1) = 150 g/mole

100 g * (1 mole/150 g) = 0.667 moles

 

5. A certain carbohydrate compound (containing only C, H and 0) is 40.0% C, 6.72% H, and 53.3% O by mass. The experimentally determined molecular mass is 180 amu. What is the empirical and chemical formula for this carbohydrate? (2 points)

Choosing 100g as a convenient mass, we would have 40.0g of C, 6.72g H and 53.3g of O

40.0 g C * (1 mole/12.0g) = 3.33 moles C

6.72 g H * (1 mole/1.01g) = 6.65 moles H

53.3 g O * (1 mole/16.0g) = 3.33 moles O

Calculating the relative stoichiometry by dividing by the smallest:

(6.65/3.33) = 2.00 moles H, and 1.0 mole each for C and O

This yields an empirical formula of CH₂O

This would have an atomic mass of (12.0 + 2.02 + 16.0) = 30 amu

The actual molecular mass is 180 amu, or (180/30) = 6.0 times that of the empirical formula. Therefore, the chemical formula must be: C₆H₁₂O₆

6. Ethanol has the chemical formula C₂H₅OH (or C₂H₆O) . If 25.0 grams of ethanol is combusted to completion, producing only CO₂ and H₂O, how many grams each of CO₂ and H₂O are produced? (2 points)

The balanced equation for the combustion would be:

C₂H₆O + 3O₂ -> 2CO₂ + 3H₂O

Thus, for one mole of ethanol combusted, the stoichiometric relationship is such that two moles of CO₂ and three moles of H₂O will be produced.

From the chemical formula ethanol has a molecular mass of 46.1 g/mole

25.0g * (1 mole/46.1g) = 0.542 moles ethanol

The stoichiometric relationships therefore are:

0.542 moles ethanol * (2 moles CO₂ / 1 mole ethanol) = 1.08 moles CO₂

0.542 moles ethanol * (3 moles H₂O / 1 mole ethanol) = 1.63 moles H₂O

The masses of CO₂ and H₂O produced are therefore:

1.08 moles CO₂ * (44.0 g/mole) = 47.5 g CO₂

1.63 moles H₂O * (18.0 g/mole) = 29.3 g H₂O

7. In the combustion of ethanol from the above example, how many grams of CO₂ would be produced if 20 g of ethanol were combusted with 20 g of O₂ ? Which is the limiting reagent? (2 points)

20.0 g ethanol * (1 mole/46.1 g) = 0.434 moles ethanol

20.0 g O₂ * (1 mole/32g) = 0.625 moles O₂

The amount of O₂ required to react completely with the given amount of ethanol:

0.434 moles ethanol * (3 moles O₂ / 1 mole ethanol) = 1.30 moles O₂

The amount of ethanol required to react completely with the given amount of O₂:

0.625 moles O₂ * (1 mole ethanol / 3 moles O₂) = 0.104 moles ethanol

Therefore, the limiting reagent is the O₂

With the given amount of O₂ we can make:

0.625 moles O₂ * (2 moles CO₂/3 moles O₂) = 0.417 moles CO₂

0.417 moles CO₂ * (44 g/mole) = 18.3 g CO₂

8. What is the molarity of a solution of LiOH made by dissolving 5.7 g of LiOH in 23.8 mls of H₂O? (1/2 point)

5.7 g LiOH * (1 mole/23.9 grams) = 0.238 moles LiOH

0.238 moles / 0.0238 liters = 10.0 moles / liter (i.e. 10.0 M)

9. You have a stock solution of 8.5 M NaOH and an unlimited amount of pure water. How would use these two to make a 0.5 liter solution of 0.5 molar NaOH? (1/2 point)