Question

Question #dd443

Answer 1

The empirical formula of the sample is `C_5H_5O`.
The molecular formula of the sample is `C_(10)H_(10)O_2`.

Explanation:

To solve this question, we need to find the amount of carbon, hydrogen and oxygen from the given data.

From 237 mg of `CO_2` we will find the amount of carbon and from 48.5 mg of `H_2O` we will find the amount of hydrogen.

After finding both, we will find the amount of oxygen by subtracting the amount of carbon and hydrogen from the total amount.

Now, let's find the mass of carbon:

237 mg of `CO_2` = 0.237 g of `CO_2`

Number of moles of `CO_2` = `frac(0.237 g)(44.01 (g)/(mol))` = 0.00539 mol

We know that 1 mol of `CO_2` contains one mole of carbon, which means the total moles of carbon in the above sample is 0.00539 mol, which is equal
to 0.0647 g of carbon.

Now, let's find the mass of hydrogen:

48.5 mg of `H_2O` = 0.0485 g of `H_2O`

Number of moles of `H_2O = frac(0.0485 g)(18.02 (g)/(mol))` = 0.00269 mol

We know that 1 mol of `H_2O` contains two moles of hydrogen, which means the total moles of hydrogen in the above sample is 0.00538 mol which is equal
to 0.00543 g of hydrogen.

Now, let's find the mass of oxygen:

We have 87.3 mg of the sample so we deduct the mass of carbon and hydrogen to find the mass of oxygen, i.e.

Mass of oxygen = 0.0873 g - 0.0647 g - 0.00543 g = 0.0172 g

Number of moles of oxygen: `frac(0.0172 g)(16.00 (g)/(mol))` = 0.00107 mol

Now, we have the composition of carbon, hydrogen and oxygen atom in the molar ratio as:

Molar ratio of C : H : O = 0.00539 : 0.00538 : 0.00107

Now, we need to find the smallest whole number ratio of molar ratio which can be done by dividing smallest number of moles.

Here the smallest number of moles is 0.00107, so divide all by 0.00107. We get:

Molar ratio of C : H : O = `frac(0.00539)(0.00107) : frac(0.00538)(0.00107) : frac(0.00107)(0.00107) = 5.04:5.03:1 ≈ 5 : 5 : 1`

Hence the empirical formula of the sample is `C_5H_5O`.

The molar mass of the empirical formula is:

`5 xx "12.01 g/mol" + 5 xx "1.008 g/mol" + 1 xx "16.00 g/mol" = "81.09 g/mol"`

Now,

Empirical formula units = `frac("molecular mass" )("empirical mass") =frac("162 g/mol")("81.09 g/mol") = 2.00 ≈ 2`

Therefore

Molecular formula = empirical formula units x empirical formula = 2 x `C_5H_5O` = `C_10H_10O_2`

Therefore the empirical formula of the sample is `C_5H_5O` and the molecular formula of the sample is `C_(10)H_(10)O_2`.