Answer :
To determine the molecular formula of a compound from its empirical formula and its molar mass, we can follow these steps:
1. Calculate the empirical formula mass (EFM):
The empirical formula of the compound is [tex]\( CH_2O \)[/tex].
- Carbon (C) has an atomic mass of 12 g/mol,
- Hydrogen (H) has an atomic mass of 1 g/mol,
- Oxygen (O) has an atomic mass of 16 g/mol.
So, the empirical formula mass is:
[tex]\[ 12 \text{ (C)} + 2 \times 1 \text{ (H)} + 16 \text{ (O)} = 12 + 2 + 16 = 30 \text{ g/mol} \][/tex]
2. Determine the molar mass:
The given molar mass of the compound is [tex]\( 180 \text{ g/mol} \)[/tex].
3. Calculate the ratio of the molar mass to the empirical formula mass:
[tex]\[ \text{Ratio} = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{180 \text{ g/mol}}{30 \text{ g/mol}} = 6 \][/tex]
4. Determine the molecular formula:
The ratio (integer factor) is 6, which tells us that the molecular formula is 6 times the empirical formula.
- Number of Carbon atoms (C) in molecular formula: [tex]\(1 \times 6 = 6\)[/tex]
- Number of Hydrogen atoms (H) in molecular formula: [tex]\(2 \times 6 = 12\)[/tex]
- Number of Oxygen atoms (O) in molecular formula: [tex]\(1 \times 6 = 6\)[/tex]
Therefore, the molecular formula is:
[tex]\[ C_6H_{12}O_6 \][/tex]
So, the correct molecular formula of the compound is [tex]\( \boxed{C_6H_{12}O_6} \)[/tex].
1. Calculate the empirical formula mass (EFM):
The empirical formula of the compound is [tex]\( CH_2O \)[/tex].
- Carbon (C) has an atomic mass of 12 g/mol,
- Hydrogen (H) has an atomic mass of 1 g/mol,
- Oxygen (O) has an atomic mass of 16 g/mol.
So, the empirical formula mass is:
[tex]\[ 12 \text{ (C)} + 2 \times 1 \text{ (H)} + 16 \text{ (O)} = 12 + 2 + 16 = 30 \text{ g/mol} \][/tex]
2. Determine the molar mass:
The given molar mass of the compound is [tex]\( 180 \text{ g/mol} \)[/tex].
3. Calculate the ratio of the molar mass to the empirical formula mass:
[tex]\[ \text{Ratio} = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{180 \text{ g/mol}}{30 \text{ g/mol}} = 6 \][/tex]
4. Determine the molecular formula:
The ratio (integer factor) is 6, which tells us that the molecular formula is 6 times the empirical formula.
- Number of Carbon atoms (C) in molecular formula: [tex]\(1 \times 6 = 6\)[/tex]
- Number of Hydrogen atoms (H) in molecular formula: [tex]\(2 \times 6 = 12\)[/tex]
- Number of Oxygen atoms (O) in molecular formula: [tex]\(1 \times 6 = 6\)[/tex]
Therefore, the molecular formula is:
[tex]\[ C_6H_{12}O_6 \][/tex]
So, the correct molecular formula of the compound is [tex]\( \boxed{C_6H_{12}O_6} \)[/tex].