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Sagot :
Let's break down the problem step by step:
### a) Elements in Hydrocarbons
Hydrocarbons are compounds made up exclusively of hydrogen (H) and carbon (C).
### b) Calculate the empirical formulae of A and B
Given data:
- Mass percent of carbon (C) in both hydrocarbons: [tex]\( 85.7\% \)[/tex]
- Molar mass of hydrocarbon A: [tex]\( 42 \)[/tex] g/mol
- Molar mass of hydrocarbon B: [tex]\( 84 \)[/tex] g/mol
- Atomic masses: Carbon ([tex]\( C \)[/tex]): [tex]\( 12 \)[/tex] g/mol, Hydrogen ([tex]\( H \)[/tex]): [tex]\( 1 \)[/tex] g/mol
#### Empirical Formula of Hydrocarbon A
1. Determine the mass of carbon in hydrocarbon A:
[tex]\[ \text{Mass of C in A} = \left( \frac{85.7}{100} \right) \times 42 = 36 \text{ g} \][/tex]
2. Determine the mass of hydrogen in hydrocarbon A:
[tex]\[ \text{Mass of H in A} = 42 - 36 = 6 \text{ g} \][/tex]
3. Convert these masses to moles:
[tex]\[ \text{Moles of C in A} = \frac{36}{12} = 3 \text{ moles} \][/tex]
[tex]\[ \text{Moles of H in A} = \frac{6}{1} = 6 \text{ moles} \][/tex]
4. Determine the simplest whole number ratio:
[tex]\[ \text{Ratio of C to H in A} = \frac{3}{3} : \frac{6}{3} = 1 : 2 \][/tex]
So, the empirical formula for hydrocarbon A is [tex]\( \text{CH}_2 \)[/tex].
#### Empirical Formula of Hydrocarbon B
1. Determine the mass of carbon in hydrocarbon B:
[tex]\[ \text{Mass of C in B} = \left( \frac{85.7}{100} \right) \times 84 = 72 \text{ g} \][/tex]
2. Determine the mass of hydrogen in hydrocarbon B:
[tex]\[ \text{Mass of H in B} = 84 - 72 = 12 \text{ g} \][/tex]
3. Convert these masses to moles:
[tex]\[ \text{Moles of C in B} = \frac{72}{12} = 6 \text{ moles} \][/tex]
[tex]\[ \text{Moles of H in B} = \frac{12}{1} = 12 \text{ moles} \][/tex]
4. Determine the simplest whole number ratio:
[tex]\[ \text{Ratio of C to H in B} = \frac{6}{6} : \frac{12}{6} = 1 : 2 \][/tex]
So, the empirical formula for hydrocarbon B is also [tex]\( \text{CH}_2 \)[/tex].
### c) Calculate the molecular formulae of A and B
Given the empirical formulae and the molar masses, let's determine the molecular formulas.
#### Molecular Formula of Hydrocarbon A
1. The empirical formula mass of [tex]\( \text{CH}_2 \)[/tex] is:
[tex]\[ 12 + (2 \times 1) = 14 \text{ g/mol} \][/tex]
2. The ratio of the molar mass to the empirical formula mass for A:
[tex]\[ \frac{42}{14} = 3 \][/tex]
3. Multiply the empirical formula by this ratio:
[tex]\[ \text{Molecular Formula of A} = (\text{CH}_2) \times 3 = \text{C}_3\text{H}_6 \][/tex]
So, the molecular formula for hydrocarbon A is [tex]\( \text{C}_3\text{H}_6 \)[/tex].
#### Molecular Formula of Hydrocarbon B
1. The empirical formula mass of [tex]\( \text{CH}_2 \)[/tex] is:
[tex]\[ 12 + (2 \times 1) = 14 \text{ g/mol} \][/tex]
2. The ratio of the molar mass to the empirical formula mass for B:
[tex]\[ \frac{84}{14} = 6 \][/tex]
3. Multiply the empirical formula by this ratio:
[tex]\[ \text{Molecular Formula of B} = (\text{CH}_2) \times 6 = \text{C}_6\text{H}_12 \][/tex]
So, the molecular formula for hydrocarbon B is [tex]\( \text{C}_6\text{H}_12 \)[/tex].
### Summary
- Elements in hydrocarbons: Carbon (C) and Hydrogen (H)
- Empirical formula of A: [tex]\( \text{CH}_2 \)[/tex]
- Empirical formula of B: [tex]\( \text{CH}_2 \)[/tex]
- Molecular formula of A: [tex]\( \text{C}_3\text{H}_6 \)[/tex]
- Molecular formula of B: [tex]\( \text{C}_6\text{H}_12 \)[/tex]
### a) Elements in Hydrocarbons
Hydrocarbons are compounds made up exclusively of hydrogen (H) and carbon (C).
### b) Calculate the empirical formulae of A and B
Given data:
- Mass percent of carbon (C) in both hydrocarbons: [tex]\( 85.7\% \)[/tex]
- Molar mass of hydrocarbon A: [tex]\( 42 \)[/tex] g/mol
- Molar mass of hydrocarbon B: [tex]\( 84 \)[/tex] g/mol
- Atomic masses: Carbon ([tex]\( C \)[/tex]): [tex]\( 12 \)[/tex] g/mol, Hydrogen ([tex]\( H \)[/tex]): [tex]\( 1 \)[/tex] g/mol
#### Empirical Formula of Hydrocarbon A
1. Determine the mass of carbon in hydrocarbon A:
[tex]\[ \text{Mass of C in A} = \left( \frac{85.7}{100} \right) \times 42 = 36 \text{ g} \][/tex]
2. Determine the mass of hydrogen in hydrocarbon A:
[tex]\[ \text{Mass of H in A} = 42 - 36 = 6 \text{ g} \][/tex]
3. Convert these masses to moles:
[tex]\[ \text{Moles of C in A} = \frac{36}{12} = 3 \text{ moles} \][/tex]
[tex]\[ \text{Moles of H in A} = \frac{6}{1} = 6 \text{ moles} \][/tex]
4. Determine the simplest whole number ratio:
[tex]\[ \text{Ratio of C to H in A} = \frac{3}{3} : \frac{6}{3} = 1 : 2 \][/tex]
So, the empirical formula for hydrocarbon A is [tex]\( \text{CH}_2 \)[/tex].
#### Empirical Formula of Hydrocarbon B
1. Determine the mass of carbon in hydrocarbon B:
[tex]\[ \text{Mass of C in B} = \left( \frac{85.7}{100} \right) \times 84 = 72 \text{ g} \][/tex]
2. Determine the mass of hydrogen in hydrocarbon B:
[tex]\[ \text{Mass of H in B} = 84 - 72 = 12 \text{ g} \][/tex]
3. Convert these masses to moles:
[tex]\[ \text{Moles of C in B} = \frac{72}{12} = 6 \text{ moles} \][/tex]
[tex]\[ \text{Moles of H in B} = \frac{12}{1} = 12 \text{ moles} \][/tex]
4. Determine the simplest whole number ratio:
[tex]\[ \text{Ratio of C to H in B} = \frac{6}{6} : \frac{12}{6} = 1 : 2 \][/tex]
So, the empirical formula for hydrocarbon B is also [tex]\( \text{CH}_2 \)[/tex].
### c) Calculate the molecular formulae of A and B
Given the empirical formulae and the molar masses, let's determine the molecular formulas.
#### Molecular Formula of Hydrocarbon A
1. The empirical formula mass of [tex]\( \text{CH}_2 \)[/tex] is:
[tex]\[ 12 + (2 \times 1) = 14 \text{ g/mol} \][/tex]
2. The ratio of the molar mass to the empirical formula mass for A:
[tex]\[ \frac{42}{14} = 3 \][/tex]
3. Multiply the empirical formula by this ratio:
[tex]\[ \text{Molecular Formula of A} = (\text{CH}_2) \times 3 = \text{C}_3\text{H}_6 \][/tex]
So, the molecular formula for hydrocarbon A is [tex]\( \text{C}_3\text{H}_6 \)[/tex].
#### Molecular Formula of Hydrocarbon B
1. The empirical formula mass of [tex]\( \text{CH}_2 \)[/tex] is:
[tex]\[ 12 + (2 \times 1) = 14 \text{ g/mol} \][/tex]
2. The ratio of the molar mass to the empirical formula mass for B:
[tex]\[ \frac{84}{14} = 6 \][/tex]
3. Multiply the empirical formula by this ratio:
[tex]\[ \text{Molecular Formula of B} = (\text{CH}_2) \times 6 = \text{C}_6\text{H}_12 \][/tex]
So, the molecular formula for hydrocarbon B is [tex]\( \text{C}_6\text{H}_12 \)[/tex].
### Summary
- Elements in hydrocarbons: Carbon (C) and Hydrogen (H)
- Empirical formula of A: [tex]\( \text{CH}_2 \)[/tex]
- Empirical formula of B: [tex]\( \text{CH}_2 \)[/tex]
- Molecular formula of A: [tex]\( \text{C}_3\text{H}_6 \)[/tex]
- Molecular formula of B: [tex]\( \text{C}_6\text{H}_12 \)[/tex]
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