Question

Elemental analysis of a pure compound indicated that the compound contained 324 g of C, 48.5 g of H and 16.0 g of O. What is its empirical formula?

Answer 1

C27H48O

Step-by-step explanation:

To determine the empirical formula of the compound, we need to find the simplest whole number ratio of atoms in the compound. We can do this by assuming that we have 100 g of the compound, and finding the number of moles of each element in this amount.

Number of moles of carbon (C): 324 g / 12.01 g/mol = 26.98 mol

Number of moles of hydrogen (H): 48.5 g / 1.01 g/mol = 48.02 mol

Number of moles of oxygen (O): 16.0 g / 16.00 g/mol = 1.00 mol

Next, we divide each of these mole values by the smallest value to get the simplest ratio:

C: 26.98 mol / 1.00 mol = 26.98

H: 48.02 mol / 1.00 mol = 48.02

O: 1.00 mol / 1.00 mol = 1.00

We can see that the simplest ratio of atoms in the compound is approximately C27H48O. However, we need to express this as a whole number ratio, so we divide each subscript by the smallest subscript (which is 1):

Empirical formula: C27H48O

Therefore, the empirical formula of the compound is C27H48O.

Answer 2

To find the empirical formula of a compound, we need to determine the simplest whole number ratio of the atoms present in the compound. We can do this by dividing each element's mass by its molar mass to get the number of moles of each element, and then dividing each number of moles by the smallest number of moles obtained. The molar masses of carbon, hydrogen, and oxygen are 12.01 g/mol, 1.008 g/mol, and 16.00 g/mol, respectively. Number of moles of C = 324 g / 12.01 g/mol = 26.98 mol Number of moles of H = 48.5 g / 1.008 g/mol = 48.11 mol Number of moles of O = 16.0 g / 16.00 g/mol = 1.