Question

What is the empirical formula for a compound if a 2.50g sample contains 0.900g of calcium and 1.60g

of chlorine? (3pts]

Answer 1

The empirical formula of the compound is CaCl2.

Step-by-step explanation:

To determine the empirical formula of a compound, you need to find the simplest whole-number ratio of the elements present in the compound. In this case, you have 2.50g of compound containing 0.900g of calcium and 1.60g of chlorine.

First, convert the mass of each element to moles by dividing the mass by the molar mass. The molar mass of calcium is 40.08 g/mol, and the molar mass of chlorine is 35.45 g/mol.

• Moles of calcium: 0.900g / 40.08 g/mol = 0.0224 mol
• Moles of chlorine: 1.60g / 35.45 g/mol = 0.0451 mol

Next, divide the moles of each element by the smallest number of moles to get the simplest whole-number ratio.

• Calcium: 0.0224 mol / 0.0224 mol = 1
• Chlorine: 0.0451 mol / 0.0224 mol = 2

The empirical formula for the compound is CaCl2, which indicates that it contains one calcium atom and two chlorine atoms.

Answer 2

CaCl₂

Step-by-step explanation:

First we convert the given masses of elements into moles, using their respective molar masses:

• 0.900 g Ca ÷ 40 g/mol = 0.0225 mol Ca

• 1.60 g Cl ÷ 35.45 g/mol = 0.045 mol Cl

Now we divide those numbers of moles by the lowest value among them:

• 0.0225 mol Ca / 0.0225 mol = 1

• 0.045 mol Cl / 0.0225 mol = 2

Meaning the empirical formula for the compound is CaCl₂.