The empirical formula of a compound with 54.04% calcium, 43.23% oxygen, and 2.70% hydrogen is found by converting percentages to moles, dividing by the smallest mole value to get whole numbers, and writing the formula based on these ratios, resulting in Ca(OH)2.
To find the empirical formula of a compound containing 54.04% calcium (Ca), 43.23% oxygen (O), and 2.70% hydrogen (H), we start by converting these percentages into moles. First, we divide each percentage by the atomic mass of each element: for Ca (atomic mass = 40.08 g/mol), O (atomic mass = 16.00 g/mol), and H (atomic mass = 1.01 g/mol). This gives us the mole ratio of the elements in the compound:
- Ca: 54.04% / 40.08 = 1.348 moles
- O: 43.23% / 16.00 = 2.702 moles
- H: 2.70% / 1.01 = 2.673 moles
We then divide each mole value by the smallest number of moles to find the simplest whole number ratio:
- Ca: 1.348 / 1.348 = 1
- O: 2.702 / 1.348 ≈ 2
- H: 2.673 / 1.348 ≈ 2
The resulting empirical formula contains one atom of calcium, two atoms of oxygen, and two atoms of hydrogen, which can be written as Ca(OH)2. This is the empirical formula for calcium hydroxide.
The empirical formula of the compound with 54.04% calcium, 43.23% oxygen, and 2.70% hydrogen is Ca(OH)2 after calculating molar ratios and simplifying to the smallest whole numbers.