29. ca(oh)₂
29. ______%ca
______%o
______%h
30. a compound contains 30.45% nitrogen and 69.55% oxygen. if its molecular mass is 92.02, what is its molecular formula?
31. octane, a compound of hydrogen and carbon found in gasoline, has a molecular mass of 114.26. if the percentage of hydrogen in octane is 15.88, what is its molecular formula?
32. a compound of phosphorus and oxygen contains 56.36% oxygen. if the molecular mass is 283.9, what is the molecular formula of the compound?
33. how many grams of oxygen are contained in a 200. gram sample of water?
34. how many grams of carbon are contained in a 250. gram sample of co₂?

Question

29. ca(oh)₂
29. ______%ca
    ______%o
    ______%h
30. a compound contains 30.45% nitrogen and 69.55% oxygen. if its molecular mass is 92.02, what is its molecular formula?
31. octane, a compound of hydrogen and carbon found in gasoline, has a molecular mass of 114.26. if the percentage of hydrogen in octane is 15.88, what is its molecular formula?
32. a compound of phosphorus and oxygen contains 56.36% oxygen. if the molecular mass is 283.9, what is the molecular formula of the compound?
33. how many grams of oxygen are contained in a 200. gram sample of water?
34. how many grams of carbon are contained in a 250. gram sample of co₂?

Accepted Answer

### Problem 29: Percent Composition of $ \text{Ca(OH)}_2 $
# Explanation:
## Step 1: Calculate Molar Mass of $ \text{Ca(OH)}_2 $
Molar mass of $ \text{Ca} = 40.08 \, \text{g/mol} $, $ \text{O} = 16.00 \, \text{g/mol} $, $ \text{H} = 1.008 \, \text{g/mol} $.
Molar mass of $ \text{Ca(OH)}_2 = 40.08 + 2 \times (16.00 + 1.008) = 40.08 + 2 \times 17.008 = 40.08 + 34.016 = 74.096 \, \text{g/mol} $.

## Step 2: Percent Composition of $ \text{Ca} $
$ \% \text{Ca} = \frac{40.08}{74.096} \times 100 \approx 54.09\% $.

## Step 3: Percent Composition of $ \text{O} $
$ \% \text{O} = \frac{2 \times 16.00}{74.096} \times 100 = \frac{32.00}{74.096} \times 100 \approx 43.19\% $.

## Step 4: Percent Composition of $ \text{H} $
$ \% \text{H} = \frac{2 \times 1.008}{74.096} \times 100 = \frac{2.016}{74.096} \times 100 \approx 2.72\% $.

# Answer:
$ \% \text{Ca} \approx 54.09\% $, $ \% \text{O} \approx 43.19\% $, $ \% \text{H} \approx 2.72\% $

### Problem 30: Molecular Formula of Compound with $ 30.45\% \, \text{N} $ and $ 69.55\% \, \text{O} $ (Molecular Mass = 92.02)
# Explanation:
## Step 1: Assume 100 g Sample
Mass of $ \text{N} = 30.45 \, \text{g} $, Mass of $ \text{O} = 69.55 \, \text{g} $.

## Step 2: Moles of Each Element
Moles of $ \text{N} = \frac{30.45}{14.01} \approx 2.173 \, \text{mol} $ (molar mass of $ \text{N} = 14.01 \, \text{g/mol} $).
Moles of $ \text{O} = \frac{69.55}{16.00} \approx 4.347 \, \text{mol} $.

## Step 3: Mole Ratio (Divide by Smallest)
Ratio $ \text{N} : \text{O} = \frac{2.173}{2.173} : \frac{4.347}{2.173} \approx 1 : 2 $. Empirical formula: $ \text{NO}_2 $.

## Step 4: Molar Mass of Empirical Formula
Molar mass of $ \text{NO}_2 = 14.01 + 2 \times 16.00 = 46.01 \, \text{g/mol} $.

## Step 5: Determine Multiplier $ n $
$ n = \frac{\text{Molecular Mass}}{\text{Empirical Mass}} = \frac{92.02}{46.01} = 2 $.

## Step 6: Molecular Formula
Molecular formula = $ (\text{NO}_2)_2 = \text{N}_2\text{O}_4 $.

# Answer:
Molecular formula: $ \text{N}_2\text{O}_4 $

### Problem 31: Molecular Formula of Octane ( $ \text{C} $ and $ \text{H} $, Molecular Mass = 114.26, $ \% \text{H} = 15.88\% $)
# Explanation:
## Step 1: Mass of Each Element (100 g Sample)
Mass of $ \text{H} = 15.88 \, \text{g} $, Mass of $ \text{C} = 100 - 15.88 = 84.12 \, \text{g} $.

## Step 2: Moles of Each Element
Moles of $ \text{H} = \frac{15.88}{1.008} \approx 15.75 \, \text{mol} $, Moles of $ \text{C} = \frac{84.12}{12.01} \approx 7.004 \, \text{mol} $.

## Step 3: Mole Ratio (Divide by Smallest)
Ratio $ \text{C} : \text{H} = \frac{7.004}{7.004} : \frac{15.75}{7.004} \approx 1 : 2.25 $. Multiply by 4 to get whole numbers: $ 4 : 9 $. Empirical formula: $ \text{C}_4\text{H}_9 $.

## Step 4: Molar Mass of Empirical Formula
Molar mass of $ \text{C}_4\text{H}_9 = 4 \times 12.01 + 9 \times 1.008 = 48.04 + 9.072 = 57.112 \, \text{g/mol} $.

## Step 5: Determine Multiplier $ n $
$ n = \frac{114.26}{57.112} \approx 2 $.

## Step 6: Molecular Formula
Molecular formula = $ (\text{C}_4\text{H}_9)_2 = \text{C}_8\text{H}_{18} $.

# Answer:
Molecular formula: $ \text{C}_8\text{H}_{18} $

### Problem 32: Molecular Formula of $ \text{P} $ and $ \text{O} $ Compound ( $ \% \text{O} = 56.36\% $, Molecular Mass = 283.9)
# Explanation:
## Step 1: Mass of Each Element (100 g Sample)
Mass of $ \text{O} = 56.36 \, \text{g} $, Mass of $ \text{P} = 100 - 56.36 = 43.64 \, \text{g} $.

## Step 2: Moles of Each Element
Moles of $ \text{P} = \frac{43.64}{30.97} \approx 1.409 \, \text{mol} $ (molar mass of $ \text{P} = 30.97 \, \text{g/mol} $).
Moles of $ \text{O} = \frac{56.36}{16.00} \approx 3.523 \, \text{mol} $.

## Step 3: Mole Ratio (Divide by Smallest)
Ratio $ \text{P} : \text{O} = \frac{1.409}{1.409} : \frac{3.523}{1.409} \approx 1 : 2.5 $. Multiply by 2 to get whole numbers: $ 2 : 5 $. Empirical formula: $ \text{P}_2\text{O}_5 $.

## Step 4: Molar Mass of Empirical Formula
Molar mass of $ \text{P}_2\text{O}_5 = 2 \times 30.97 + 5 \times 16.00 = 61.94 + 80.00 = 141.94 \, \text{g/mol} $.

## Step 5: Determine Multiplier $ n $
$ n = \frac{283.9}{141.94} \approx 2 $.

## Step 6: Molecular Formula
Molecular formula = $ (\text{P}_2\text{O}_5)_2 = \text{P}_4\text{O}_{10} $.

# Answer:
Molecular formula: $ \text{P}_4\text{O}_{10} $

### Problem 33: Grams of $ \text{O} $ in 200 g $ \text{H}_2\text{O} $
# Explanation:
## Step 1: Molar Mass of $ \text{H}_2\text{O} $
Molar mass of $ \text{H}_2\text{O} = 2 \times 1.008 + 16.00 = 18.016 \, \text{g/mol} $.

## Step 2: Mass Fraction of $ \text{O} $ in $ \text{H}_2\text{O} $
Mass fraction of $ \text{O} = \frac{16.00}{18.016} \approx 0.8881 $.

## Step 3: Grams of $ \text{O} $
Grams of $ \text{O} = 200 \, \text{g} \times 0.8881 \approx 177.6 \, \text{g} $.

# Answer:
Grams of $ \text{O} \approx 177.6 \, \text{g} $

### Problem 34: Grams of $ \text{C} $ in 250 g $ \text{CO}_2 $
# Explanation:
## Step 1: Molar Mass of $ \text{CO}_2 $
Molar mass of $ \text{CO}_2 = 12.01 + 2 \times 16.00 = 44.01 \, \text{g/mol} $.

## Step 2: Mass Fraction of $ \text{C} $ in $ \text{CO}_2 $
Mass fraction of $ \text{C} = \frac{12.01}{44.01} \approx 0.2729 $.

## Step 3: Grams of $ \text{C} $
Grams of $ \text{C} = 250 \, \text{g} \times 0.2729 \approx 68.2 \, \text{g} $.

# Answer:
Grams of $ \text{C} \approx 68.2 \, \text{g} $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 29: Percent Composition of \( \text{Ca(OH)}_2 \)

Step 1: Calculate Molar Mass of \( \text{Ca(OH)}_2 \)

Step 2: Percent Composition of \( \text{Ca} \)

Step 3: Percent Composition of \( \text{O} \)

Step 4: Percent Composition of \( \text{H} \)

Step 1: Assume 100 g Sample

Step 2: Moles of Each Element

Step 3: Mole Ratio (Divide by Smallest)

Step 4: Molar Mass of Empirical Formula

Step 5: Determine Multiplier \( n \)

Step 6: Molecular Formula

Step 1: Mass of Each Element (100 g Sample)

Step 2: Moles of Each Element

Step 3: Mole Ratio (Divide by Smallest)

Step 4: Molar Mass of Empirical Formula

Step 5: Determine Multiplier \( n \)

Step 6: Molecular Formula

Answer:

Problem 30: Molecular Formula of Compound with \( 30.45\% \, \text{N} \) and \( 69.55\% \, \text{O} \) (Molecular Mass = 92.02)