Question

Determine the empirical formula of a compound containing 48.38 grams of carbon, 6.74 grams of hydrogen, and 53.5 grams of oxygen.

In an experiment, the molar mass of the compound was determined to be 180.15 g/mol. What is the molecular formula of the compound?

For both questions, show your work or explain how you determined the formulas by giving specific values used in calculations.

Answer 1

First, I will convert from grams to moles. This will give me approximately 4.032 moles of carbon, 6.740 moles of hydrogen, and 3.344 moles of oxygen. Then, I will calculate the mole ratio, using whole numbers. I must also identify the least amount of moles in an element for this step, which is oxygen at 3.344 moles.

Oxygen = 3.344/3.344 = 1

Carbon = 4.032/3.344 = 1

Hydrogen = 6.740/3.344 = 2

The empirical formula will be COH2.

Now that we have the molar mass of the molecular formula, we can calculate for the molecular formula. First, we will begin by discovering the molar mass of the empirical formula, which will be C molar mass x 1 + O molar mass x 1 + H molar mass x 2. That will be equal to 60.949. After that, we can divide the molecular mass by the empirical formula molar mass, giving us approximately 3. Now we will multiply the empirical formula by 3, giving us our molecular formula, which will be equal to C3O3H6.

Answer 2

Hence, C121H199O100 represents the compound's empirical formula.

Step-by-step explanation:

We must ascertain the ratio of the number of atoms of each element in the compound in order to derive the empirical formula. To accomplish this, we can divide the mole ratio by the least number of moles after converting the masses of each element to moles.

The elements' molar masses are as follows:

12.01 g/mol for carbon

1.01 g/mol for hydrogen

16.00 g/mol for oxygen

When we convert the masses to moles, we obtain:

48.38 g / 12.01 g/mol = 4.03 mol of carbon

6.74 g / 1.01 g/mol of hydrogen equals 6.67 mol

53.5 g / 16.00 g/mol = 3.34 mol for oxygen

3.34 mol of oxygen is the least amount of moles. If you divide the total moles of each element by 3.34 mol, you get:

Carbon: 4.03 mol/3.3 mol = 1.21 mol Hydrogen:6.67 moles/3.34 moles equals 1.99

3.34 mol / 3.34 mol = 1.00 for oxygen.

By multiplying each ratio by 100 to obtain whole integers, we may simplify the ratios, which are around 1.21:1.99:1.00. This results in a ratio of roughly 121:199:100.

We divide each number in the ratio by their greatest common factor (GCF) to arrive at the empirical formula. 121, 199, and 100 have a GCF of 1. By dividing by 1, we get:

Carbohydrate: 121 / 1 = 121

199 / 1 = 199 for hydrogen.

100 / 1 Equals 100 for oxygen.

Hence, C121H199O100 represents the compound's empirical formula.