Determine the empirical formula of a compound containing 40.6 grams of carbon, 5.1 grams of hydrogen, and 54.2 grams of oxygen. In an experiment, the molar mass of the compound was determined to be 118.084 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.

Empirical formula is C2H3O2

Explanation;

The empirical formula of a compound represents the simplest whole number ratio of elements in the compound.

First we determine the moles of each element in the compound by dividing the given mass by the molar mass.

Carbon; 40.6 g/12 g/mol = 3.3833 moles

Hydrogen; 5.1 g/1 g/mol = 5.1 moles

Oxygen; 54.2 g/16 g/mol = 3.3875 moles

We then determine the mole ratio; by dividing with the smallest number of moles;

C = 3.3833/3.3833 = 1.00

H = 5.1 /3.3833 =    1.5

O =3.3875/3.3833  = 1.00

The ratio = 1 : 1.5 : 1, we multiply to make 1.5 a whole number;

we get, 2: 3: 2

Therefore; the empirical formula = C2H3O2

2. Answer;

Molecular formula =C4H6O4

Explanation;

Molecular formula shows the number of atoms of each element in a compound.

Thus; (C2H3O2)n = 118.084

= (24 + 3 +32)n = 118.084

= 59n =118.084

n = 2.00

Therefore, molecular formula = (C2H3O2)2

                                                = C4H6O4

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.

im not exactly sure but i think its b!