Answer :
To find the molecular formulae for the given compounds, we need to follow these steps:
1. Calculate the empirical mass of each compound using the molar masses of the elements provided.
2. Determine the multiplier by dividing the molar mass [tex]\( M_r \)[/tex] of the compound by the empirical mass.
3. Multiply the subscripts in the empirical formula by this multiplier to get the molecular formula.
Let's do these step by step for each compound:
### (a) Hydrazine
- Empirical formula: [tex]\( NH_2 \)[/tex]
- Molar mass ( [tex]\( M_r \)[/tex] ): 32
- Molar masses: N = 14, H = 1
1. Calculate empirical mass:
[tex]\( \text{Empirical mass} = 1 \times 14 + 2 \times 1 = 14 + 2 = 16 \)[/tex]
2. Determine the multiplier:
[tex]\( \text{Multiplier} = \frac{32}{16} = 2 \)[/tex]
3. Molecular formula:
Multiply the subscripts in [tex]\( NH_2 \)[/tex] by 2:
[tex]\( N_{2 \times 1}H_{2 \times 2} = N_2H_4 \)[/tex]
Therefore, the molecular formula for hydrazine is [tex]\( N_2H_4 \)[/tex].
### (b) Cyanogen
- Empirical formula: [tex]\( CN \)[/tex]
- Molar mass ( [tex]\( M_r \)[/tex] ): 52
- Molar masses: C = 12, N = 14
1. Calculate empirical mass:
[tex]\( \text{Empirical mass} = 1 \times 12 + 1 \times 14 = 12 + 14 = 26 \)[/tex]
2. Determine the multiplier:
[tex]\( \text{Multiplier} = \frac{52}{26} = 2 \)[/tex]
3. Molecular formula:
Multiply the subscripts in [tex]\( CN \)[/tex] by 2:
[tex]\( C_{2 \times 1}N_{2 \times 1} = C_2N_2 \)[/tex]
Therefore, the molecular formula for cyanogen is [tex]\( C_2N_2 \)[/tex].
### (c) Nitrogen Oxide
- Empirical formula: [tex]\( NO_2 \)[/tex]
- Molar mass ( [tex]\( M_r \)[/tex] ): 92
- Molar masses: N = 14, O = 16
1. Calculate empirical mass:
[tex]\( \text{Empirical mass} = 1 \times 14 + 2 \times 16 = 14 + 32 = 46 \)[/tex]
2. Determine the multiplier:
[tex]\( \text{Multiplier} = \frac{92}{46} = 2 \)[/tex]
3. Molecular formula:
Multiply the subscripts in [tex]\( NO_2 \)[/tex] by 2:
[tex]\( N_{2 \times 1}O_{2 \times 2} = N_2O_4 \)[/tex]
Therefore, the molecular formula for nitrogen oxide is [tex]\( N_2O_4 \)[/tex].
### (d) Glucose
- Empirical formula: [tex]\( CH_2O \)[/tex]
- Molar mass ( [tex]\( M_r \)[/tex] ): 180
- Molar masses: C = 12, H = 1, O = 16
1. Calculate empirical mass:
[tex]\( \text{Empirical mass} = 1 \times 12 + 2 \times 1 + 1 \times 16 = 12 + 2 + 16 = 30 \)[/tex]
2. Determine the multiplier:
[tex]\( \text{Multiplier} = \frac{180}{30} = 6 \)[/tex]
3. Molecular formula:
Multiply the subscripts in [tex]\( CH_2O \)[/tex] by 6:
[tex]\( C_{6 \times 1}H_{6 \times 2}O_{6 \times 1} = C_6H_{12}O_6 \)[/tex]
Therefore, the molecular formula for glucose is [tex]\( C_6H_{12}O_6 \)[/tex].
### Summary
| Compound | [tex]\( M_r \)[/tex] | Empirical Formula | Molecular Formula |
|---------------|---------|-------------------|--------------------|
| Hydrazine | 32 | [tex]\( NH_2 \)[/tex] | [tex]\( N_2H_4 \)[/tex] |
| Cyanogen | 52 | [tex]\( CN \)[/tex] | [tex]\( C_2N_2 \)[/tex] |
| Nitrogen Oxide| 92 | [tex]\( NO_2 \)[/tex] | [tex]\( N_2O_4 \)[/tex] |
| Glucose | 180 | [tex]\( CH_2O \)[/tex] | [tex]\( C_6H_{12}O_6 \)[/tex] |
1. Calculate the empirical mass of each compound using the molar masses of the elements provided.
2. Determine the multiplier by dividing the molar mass [tex]\( M_r \)[/tex] of the compound by the empirical mass.
3. Multiply the subscripts in the empirical formula by this multiplier to get the molecular formula.
Let's do these step by step for each compound:
### (a) Hydrazine
- Empirical formula: [tex]\( NH_2 \)[/tex]
- Molar mass ( [tex]\( M_r \)[/tex] ): 32
- Molar masses: N = 14, H = 1
1. Calculate empirical mass:
[tex]\( \text{Empirical mass} = 1 \times 14 + 2 \times 1 = 14 + 2 = 16 \)[/tex]
2. Determine the multiplier:
[tex]\( \text{Multiplier} = \frac{32}{16} = 2 \)[/tex]
3. Molecular formula:
Multiply the subscripts in [tex]\( NH_2 \)[/tex] by 2:
[tex]\( N_{2 \times 1}H_{2 \times 2} = N_2H_4 \)[/tex]
Therefore, the molecular formula for hydrazine is [tex]\( N_2H_4 \)[/tex].
### (b) Cyanogen
- Empirical formula: [tex]\( CN \)[/tex]
- Molar mass ( [tex]\( M_r \)[/tex] ): 52
- Molar masses: C = 12, N = 14
1. Calculate empirical mass:
[tex]\( \text{Empirical mass} = 1 \times 12 + 1 \times 14 = 12 + 14 = 26 \)[/tex]
2. Determine the multiplier:
[tex]\( \text{Multiplier} = \frac{52}{26} = 2 \)[/tex]
3. Molecular formula:
Multiply the subscripts in [tex]\( CN \)[/tex] by 2:
[tex]\( C_{2 \times 1}N_{2 \times 1} = C_2N_2 \)[/tex]
Therefore, the molecular formula for cyanogen is [tex]\( C_2N_2 \)[/tex].
### (c) Nitrogen Oxide
- Empirical formula: [tex]\( NO_2 \)[/tex]
- Molar mass ( [tex]\( M_r \)[/tex] ): 92
- Molar masses: N = 14, O = 16
1. Calculate empirical mass:
[tex]\( \text{Empirical mass} = 1 \times 14 + 2 \times 16 = 14 + 32 = 46 \)[/tex]
2. Determine the multiplier:
[tex]\( \text{Multiplier} = \frac{92}{46} = 2 \)[/tex]
3. Molecular formula:
Multiply the subscripts in [tex]\( NO_2 \)[/tex] by 2:
[tex]\( N_{2 \times 1}O_{2 \times 2} = N_2O_4 \)[/tex]
Therefore, the molecular formula for nitrogen oxide is [tex]\( N_2O_4 \)[/tex].
### (d) Glucose
- Empirical formula: [tex]\( CH_2O \)[/tex]
- Molar mass ( [tex]\( M_r \)[/tex] ): 180
- Molar masses: C = 12, H = 1, O = 16
1. Calculate empirical mass:
[tex]\( \text{Empirical mass} = 1 \times 12 + 2 \times 1 + 1 \times 16 = 12 + 2 + 16 = 30 \)[/tex]
2. Determine the multiplier:
[tex]\( \text{Multiplier} = \frac{180}{30} = 6 \)[/tex]
3. Molecular formula:
Multiply the subscripts in [tex]\( CH_2O \)[/tex] by 6:
[tex]\( C_{6 \times 1}H_{6 \times 2}O_{6 \times 1} = C_6H_{12}O_6 \)[/tex]
Therefore, the molecular formula for glucose is [tex]\( C_6H_{12}O_6 \)[/tex].
### Summary
| Compound | [tex]\( M_r \)[/tex] | Empirical Formula | Molecular Formula |
|---------------|---------|-------------------|--------------------|
| Hydrazine | 32 | [tex]\( NH_2 \)[/tex] | [tex]\( N_2H_4 \)[/tex] |
| Cyanogen | 52 | [tex]\( CN \)[/tex] | [tex]\( C_2N_2 \)[/tex] |
| Nitrogen Oxide| 92 | [tex]\( NO_2 \)[/tex] | [tex]\( N_2O_4 \)[/tex] |
| Glucose | 180 | [tex]\( CH_2O \)[/tex] | [tex]\( C_6H_{12}O_6 \)[/tex] |
