<span>Calculating the moles and the moles ratio of the elements gives us the ratio of atoms in each element.
Converting the percentage of element into grams
40.25% carbon = 40.25/100 = .4025 * 100 g of carbon = 40.25g of C
6.19% hydrogen = 6.19/100 = .0619 * 100g g of hydrogen = 6.19g of H
8.94% oxygen = 8.94/100 = .0819 * 100 g of oxygen = 8.19g of O
44.62% bromine = 44.62/100 = .4462 * 100 g of bromine = 44.62g of Br
Converting the grams of element into moles
(48.38 g C) (1 mol/ 12.10 g C) = 4.028 mol C
(8.12 g H) (1 mol/ 1.008 g H) = 8.056 mol H
(53.38 g O) (1 mol/ 16.00 g O) = 3.336 mol O
(44.62g of Br)(0.012515018021626 moles) = 0.55842 mol Br
Calculating the moles ratio of elements by dividing the small number of moles of an element
4.028 mol C /0.55842 = 7.2 mol C x 5 = 36 mol C
8.056 mol H / 0.55842 = 14.42 mol H = 72 mol H
3.336 mol O / 0.55842 = 5.97 mol O = 30 mol O
0.55842 mol Br / 0.55842 = 1mol Br = 5 mol Br
So the empirical formula is (C6H12O5)6Br5</span>
Molarity is defined as the number of moles of solute in 1 L of solution molarity of stock solution to be prepared - 100 x 10⁻³ mol/L volume of stock solution to be prepared - 1.2 mL Therefore number of moles in 1.2 mL - 100 x 10⁻³ mol/L x 1.2 x 10⁻³ L number of moles of drug - 1.2 x 10⁻⁴ mol mass of drug required - 1.2 x 10⁻⁴ mol x 181.6 g/mol = 21. 8 mg 21.8 g of drug is required to make the stock solution
