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# A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. If the molar m...

### Question

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. If the molar mass of the compound is 180. Find the molecular formula.

[H = 1, C = 12, O = 16]

### Options

A)
C$$_3$$H$$_6$$O$$_3$$
B)
C$$_6$$H$$_6$$O$$_3$$
C)
C$$_6$$H$$_{12}$$O$$_6$$
D)
CH$$_2$$O

### Explanation:

To find the molecular formula of the compound, first, we need to determine the empirical formula. This can be done by dividing the percentage of each element by its atomic mass and then finding the ratio of the resulting values.

For carbon (C): $$\frac{40.0\%}{12} = 3.33$$

For hydrogen (H): $$\frac{6.7\%}{1} = 6.7$$

For oxygen (O): $$\frac{53.3\%}{16} = 3.33$$

Next, find the simplest whole-number ratio by dividing each value by the smallest value, which in this case is 3.33:

For carbon: $$\frac{3.33}{3.33} = 1$$

For hydrogen: $$\frac{6.7}{3.33} \approx 2$$

For oxygen: $$\frac{3.33}{3.33} = 1$$

So, the empirical formula is CH$$_2$$O.

Now, we need to find the molecular formula using the given molar mass of the compound, which is 180. First, calculate the molar mass of the empirical formula:

Molar mass of CH$$_2$$O = (1 × 12) + (2 × 1) + (1 × 16) = 12 + 2 + 16 = 30

To find the molecular formula, divide the given molar mass by the molar mass of the empirical formula and multiply the empirical formula by the result:

$$\frac{180}{30} = 6$$

So, the molecular formula is $$6 \times$$CH$$_2$$O = C$$_6$$H$$_{12}$$O$$_6$$, which is Option C.