Unit 1: Atomic Structure and Properties
Essential Knowledge
◼ Some pure substances are composed of individual molecules, while others consist of atoms or ions held together in fixed proportions as described by a formula unit. (SPQ-2.A.1)
◼ According to the law of definite proportions, the ratio of the masses of the constituent elements in any pure sample of that compound is always the same. (SPQ-2.A.2)
◼ The chemical formula that lists the lowest whole number ratio of atoms of the elements in a compound is the empirical formula. (SPQ-2.A.3)
◼ Some pure substances are composed of individual molecules, while others consist of atoms or ions held together in fixed proportions as described by a formula unit. (SPQ-2.A.1)
Describe the particle (i.e., atom, molecule, ion, formula unit) the makes up the followings:
Sodium chloride
Water
Graphite (carbon)
Calcium chloride
Sodium chloride is composed of sodium ions and chloride ions in a 1:1 ratio and the formula unit is represented as NaCl
Water is a molecule that is comprised of two hydrogen atoms covalently bonded to one oxygen atom.
Graphite is carbon atoms that are covalent network - there are no individual molecules
Calcium chloride is comprised of calcium and chloride ions in a 1:2 ratio and the formula unit is represented as CaCl2
Try it yourself …
Describe the particle (i.e., atom, molecule, ion, formula unit) the makes up the followings:
a) Potassium bromide (KBr)
b) Carbon dioxide (CO₂)
c) Diamond (carbon)
d) Magnesium oxide (MgO)
Check your answer here.
◼ According to the law of definite proportions, the ratio of the masses of the constituent elements in any pure sample of that compound is always the same. (SPQ-2.A.2)
◼ The chemical formula that lists the lowest whole number ratio of atoms of the elements in a compound is the empirical formula. (SPQ-2.A.3)
Iron can form three different oxides, FeO, Fe2O3 and Fe3O4. A sample of iron oxide was analyzed and was found to contain 69.943% iron with the rest of the mass from oxygen.
(a) Determine the empirical formula to determine the identity of the iron oxide.
(b) Another sample of iron oxide was collected and it was determined to have 1.498 g of iron and 0.6437 g of oxygen.
Part (a): Determine the Empirical Formula
1. Calculate the mass percentages:
Given: 69.943% iron (Fe)
Mass of iron: 69.943 g
Mass of oxygen: 100 g - 69.943 g = 30.057 g
2. Convert grams to moles:
Convert Mass to Moles:
Moles of Fe = $\frac{69.943 \, \text{g}}{55.845 \, \frac{\text{g}}{\text{mol}}}$ = 1.252 mol
Moles of O = $\frac{30.057 \, \text{g}}{16.00 \, \frac{\text{g}}{\text{mol}}} $= 1.879 mol
3. Determine the mole ratio by dividing by the smallest number of moles
$\text{Ratio of Fe} = \frac{1.252}{1.252} = 1 $
$\text{Ratio of O} = \frac{1.879}{1.252} = 1.501 \approx 1.5 $
4. Adjust the ratio to the nearest whole numbers:
The ratio 1:1.50 can be multiplied by 2 to get whole numbers:
$1 \times 2 = 2 , (\text{Fe}) $
$1.5 \times 2 = 3 , (\text{O}) $
Therefore the empirical formula is $\text{Fe}_2\text{O}_3$
Part (b): Empirical Formula from Masses
2. Convert the given grams to moles:
For iron (Fe):
moles of Fe = $\frac{1.498 \, \text{g}}{55.85 \, \frac{\text{g}}{\text{mol}}}$ = 0.0268 mol
For oxygen (O):
$\text{moles of O} = \frac{0.6437 \, \text{g}}{16.00 \, \frac{\text{g}}{\text{mol}}} = 0.0402 \, \text{mol}$
3.Determine the mole ratio by dividing by the smallest number of moles:
$\text{ratio of Fe to O} = \frac{\text{moles of Fe}}{\text{smallest number of moles}} \, \text{and} \, \frac{\text{moles of O}}{\text{smallest number of moles}}$
$\text{ratio of Fe} = \frac{0.0268 \, \text{mol}}{0.0268 \, \text{mol}} = 1$
$\text{ratio of O} = \frac{0.0402 \, \text{mol}}{0.0268 \, \text{mol}}$ = 1.5$
Adjust the ratio to the nearest whole numbers. The ratio 1 : 1.50 can be multiplied by 2 to get whole numbers:
$\text{ratio of Fe} = 1 \times 2 = 2$
$\text{ratio of O} = 1.5 \times 2 = 3$
Answer: Therefore the empirical formula is Fe2O3
Try it yourself …
A compound known as aluminum sulfide (Al₂S₃) is composed of aluminum and sulfur. Two different samples of aluminum sulfide are analyzed. The first sample weighs 15.00 grams and is found to contain 7.93 grams of aluminum. The second sample weighs 22.50 grams and contains 11.90 grams of aluminum.
Part A: Calculate the mass percent of aluminum and sulfur in each sample.
Part B: Verify whether the Law of Definite Proportions is observed in these samples by showing that the ratio of the masses of sulfur to aluminum is constant.
Part C: Given that the molar mass of aluminum is 26.98 g/mol and sulfur is 32.06 g/mol, calculate the empirical formula of aluminum sulfide using the data from the first sample. Show your work.
Show all your calculations and explain your reasoning.