1) The vanillin molecule is the primary molecule present in vanilla extract. The molecular mass of vanillin is 152.08 grams per mole and contains 63.18% carbon, 5.26% hydrogen, and 31.56% oxygen. What is the molecular formula of vanillin?2) A sample of fuel is found to contain 87.4% nitrogen and 12.6% hydrogen. If the molecular mass of the fuel is 32.05 grams/mole, what is the molecular formula of the fuel?3) A compound is analyzed and found to have a mass of 289.9 grams/mole and contain 49.67% carbon, 48.92% chlorine and 1.39% hydrogen. What is the molecular formula of the compound?4) A compound is analyzed and found to contain 35.66% carbon, 16.24% hydrogen and 45.10% nitrogen. What is the empirical formula of the compound?
The Correct Answer and Explanation is :
Let’s break down each problem and calculate the molecular formulas step by step.
1) Molecular formula of vanillin:
Vanillin’s molecular mass is given as 152.08 g/mol, and it is made up of 63.18% carbon (C), 5.26% hydrogen (H), and 31.56% oxygen (O). We will use the percentages to find the empirical formula and then the molecular formula.
Step 1: Find the mass of each element in one mole of vanillin.
- Carbon: ( 63.18\% \times 152.08 \, \text{g} = 95.98 \, \text{g of C} )
- Hydrogen: ( 5.26\% \times 152.08 \, \text{g} = 8.00 \, \text{g of H} )
- Oxygen: ( 31.56\% \times 152.08 \, \text{g} = 48.04 \, \text{g of O} )
Step 2: Convert grams to moles.
- Moles of carbon: ( \frac{95.98 \, \text{g}}{12.01 \, \text{g/mol}} = 7.99 \, \text{mol} )
- Moles of hydrogen: ( \frac{8.00 \, \text{g}}{1.008 \, \text{g/mol}} = 7.94 \, \text{mol} )
- Moles of oxygen: ( \frac{48.04 \, \text{g}}{16.00 \, \text{g/mol}} = 3.00 \, \text{mol} )
Step 3: Determine the ratio of moles.
- Carbon: ( \frac{7.99}{3.00} = 2.66 )
- Hydrogen: ( \frac{7.94}{3.00} = 2.65 )
- Oxygen: ( \frac{3.00}{3.00} = 1 )
The ratio of C:H:O is approximately 2.66:2.65:1, which simplifies to 3:3:1. The empirical formula is ( \text{C}_8 \text{H}_8 \text{O} ).
Step 4: Compare the empirical formula mass with the molecular mass.
- Empirical formula mass = ( 8 \times 12.01 + 8 \times 1.008 + 16.00 = 152.08 \, \text{g/mol} )
Since the empirical formula mass is equal to the molecular mass, the molecular formula of vanillin is ( \text{C}_8 \text{H}_8 \text{O} ).
2) Molecular formula of the fuel:
The fuel contains 87.4% nitrogen (N) and 12.6% hydrogen (H), and its molecular mass is 32.05 g/mol.
Step 1: Find the mass of each element in one mole of the fuel.
- Nitrogen: ( 87.4\% \times 32.05 \, \text{g} = 27.99 \, \text{g of N} )
- Hydrogen: ( 12.6\% \times 32.05 \, \text{g} = 4.03 \, \text{g of H} )
Step 2: Convert grams to moles.
- Moles of nitrogen: ( \frac{27.99 \, \text{g}}{14.01 \, \text{g/mol}} = 2.00 \, \text{mol} )
- Moles of hydrogen: ( \frac{4.03 \, \text{g}}{1.008 \, \text{g/mol}} = 4.00 \, \text{mol} )
Step 3: Determine the ratio of moles.
- Nitrogen: ( \frac{2.00}{2.00} = 1 )
- Hydrogen: ( \frac{4.00}{2.00} = 2 )
The ratio of N:H is 1:2. Thus, the empirical formula is ( \text{NH}_2 ).
Step 4: Compare the empirical formula mass with the molecular mass.
- Empirical formula mass = ( 14.01 + 2 \times 1.008 = 16.03 \, \text{g/mol} )
The ratio of the molecular mass to the empirical formula mass is ( \frac{32.05}{16.03} = 2 ). Thus, the molecular formula is ( \text{N}_2 \text{H}_4 ).
3) Molecular formula of the compound:
The compound has a molecular mass of 289.9 g/mol and contains 49.67% carbon, 48.92% chlorine, and 1.39% hydrogen.
Step 1: Find the mass of each element in one mole of the compound.
- Carbon: ( 49.67\% \times 289.9 \, \text{g} = 144.92 \, \text{g of C} )
- Chlorine: ( 48.92\% \times 289.9 \, \text{g} = 141.89 \, \text{g of Cl} )
- Hydrogen: ( 1.39\% \times 289.9 \, \text{g} = 4.03 \, \text{g of H} )
Step 2: Convert grams to moles.
- Moles of carbon: ( \frac{144.92 \, \text{g}}{12.01 \, \text{g/mol}} = 12.08 \, \text{mol} )
- Moles of chlorine: ( \frac{141.89 \, \text{g}}{35.45 \, \text{g/mol}} = 4.00 \, \text{mol} )
- Moles of hydrogen: ( \frac{4.03 \, \text{g}}{1.008 \, \text{g/mol}} = 4.00 \, \text{mol} )
Step 3: Determine the ratio of moles.
- Carbon: ( \frac{12.08}{4.00} = 3.02 )
- Chlorine: ( \frac{4.00}{4.00} = 1 )
- Hydrogen: ( \frac{4.00}{4.00} = 1 )
The ratio of C:Cl:H is approximately 3:1:1. The empirical formula is ( \text{C}_3 \text{H}_3 \text{Cl} ).
Step 4: Compare the empirical formula mass with the molecular mass.
- Empirical formula mass = ( 3 \times 12.01 + 3 \times 1.008 + 35.45 = 106.29 \, \text{g/mol} )
The ratio of the molecular mass to the empirical formula mass is ( \frac{289.9}{106.29} \approx 2.73 ). This suggests that the molecular formula is approximately ( \text{C}_9 \text{H}_9 \text{Cl}_3 ).
4) Empirical formula of the compound:
The compound contains 35.66% carbon, 16.24% hydrogen, and 45.10% nitrogen.
Step 1: Find the mass of each element in one mole of the compound.
- Carbon: ( 35.66\% \times 100 \, \text{g} = 35.66 \, \text{g of C} )
- Hydrogen: ( 16.24\% \times 100 \, \text{g} = 16.24 \, \text{g of H} )
- Nitrogen: ( 45.10\% \times 100 \, \text{g} = 45.10 \, \text{g of N} )
Step 2: Convert grams to moles.
- Moles of carbon: ( \frac{35.66 \, \text{g}}{12.01 \, \text{g/mol}} = 2.97 \, \text{mol} )
- Moles of hydrogen: ( \frac{16.24 \, \text{g}}{1.008 \, \text{g/mol}} = 16.12 \, \text{mol} )
- Moles of nitrogen: ( \frac{45.10 \, \text{g}}{14.01 \, \text{g/mol}} = 3.22 \, \text{mol} )
Step 3: Determine the ratio of moles.
- Carbon: ( \frac{2.97}{2.97} = 1 )
- Hydrogen: ( \frac{16.12}{2.97} = 5.43 )
- Nitrogen: ( \frac{3.22}{2.97} = 1.08 )
The ratio of C:H:N is approximately 1:5:1, which rounds to an empirical formula of ( \text{CH}_5 \text{N} ). However, this seems unusual as ( \text{H}_5 ) is rare, so further analysis of empirical formulas may be needed.