Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
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]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
