Question

An analytical chemist analyzed a sample of a compound that is found to contain 31.25% Calcium, 18.75% Carbon and 50.0% Oxygen. What is the empirical formula of the compound? How about the molecular formula? The molar mass is 128 g/mol

Answer 1

Answer:

the empirical formula of the compound = CaC₂O₄

the molecular formula of the compound = CaC₂O₄ (calcium oxalate)

Explanation:

Step 1:

Carbon: Percentage of carbon = 18.75 %

We know that the atomic mass of carbon = 12$g/mol$

in 100 grams of the compound number of moles of carbon atoms = $\frac{18.75}{12}$

                                                                                       = 1.56 moles

Calcium: percentage of calcium = 31.25%

We know that the atomic mass of calcium = 40$g/mol$

in 100 grams of the compound number of moles of calcium atoms = $\frac{31.25}{40}$

                                                                                      = 0.78 moles

Oxygen: Percentage of oxygen = 50%

we know that the atomic mass of oxygen = 16$g/mol$

in 100 grams of the compound number of moles of oxygen atoms = $\frac{50}{16}$

                                                                                       = 3.12 moles

Here the smallest proportion is of calcium which is 0.78 moles

divide all other elements' mole numbers by the smallest one to get the whole number ratio of atoms in the compound.

carbon ⇒ $\frac{1.56}{0.78}$ = 2

calcium⇒$\frac{0.78}{0.78}$ = 1

Oxygen⇒$\frac{3.12}{0.78}$ = 4

Hence the simplest whole-number ratio = Ca:C:O = 1:2:4

empirical formula of the compound = CaC₂O₄

Empirical formula mass = 40+(12×2)+(16×4)

                                       = 128 $g/mol$

Here empirical formula mass is the same with molar mass hence empirical and molecular formula is also the same.