Certainly! Here are six problems involving empirical formulas along with their solutions:
Problem 1:
A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula.
Solution 1:
To find the empirical formula, we need to convert the percentages to moles and then find the simplest whole-number ratio.
Convert the percentages to moles:
Carbon (C): 40% = 40 grams
Hydrogen (H): 6.7% = 6.7 grams
Oxygen (O): 53.3% = 53.3 grams
Calculate the moles of each element:
Moles of C = 40 g / 12.01 g/mol (molar mass of C) ≈ 3.33 moles
Moles of H = 6.7 g / 1.01 g/mol (molar mass of H) ≈ 6.63 moles
Moles of O = 53.3 g / 16.00 g/mol (molar mass of O) ≈ 3.33 moles
Find the simplest whole number ratio by dividing all mole values by the smallest mole value (3.33 moles):
C:H:O = 3.33 moles : 6.63 moles : 3.33 moles
Simplifying by dividing by 3.33: C:H:O = 1:2:1
So, the empirical formula is CH2O.
Problem 2:
A compound contains 28.1% sodium, 2.2% sulfur, and 69.7% oxygen by mass. Determine its empirical formula.
Solution 2:
Follow the same steps as in Problem 1:
Convert the percentages to moles:
Sodium (Na): 28.1% = 28.1 grams
Sulfur (S): 2.2% = 2.2 grams
Oxygen (O): 69.7% = 69.7 grams
Calculate the moles of each element:
Moles of Na = 28.1 g / 22.99 g/mol (molar mass of Na) ≈ 1.22 moles
Moles of S = 2.2 g / 32.07 g/mol (molar mass of S) ≈ 0.07 moles
Moles of O = 69.7 g / 16.00 g/mol (molar mass of O) ≈ 4.36 moles
Find the simplest whole number ratio by dividing all mole values by the smallest mole value (0.07 moles):
Na:S:O = 1.22 moles : 0.07 moles : 4.36 moles
Simplifying by dividing by 0.07: Na:S:O = 17.43 moles : 1 moles : 62.29 moles
Rounding to the nearest whole numbers: Na:S:O ≈ 17:1:62
So, the empirical formula is Na2S62O.
Problem 3:
A compound is found to contain 29.3% phosphorus and 70.7% oxygen by mass. Determine its empirical formula.
Solution 3:
Follow the same steps as in Problem 1:
Convert the percentages to moles:
Phosphorus (P): 29.3% = 29.3 grams
Oxygen (O): 70.7% = 70.7 grams
Calculate the moles of each element:
Moles of P = 29.3 g / 30.97 g/mol (molar mass of P) ≈ 0.95 moles
Moles of O = 70.7 g / 16.00 g/mol (molar mass of O) ≈ 4.42 moles
Find the simplest whole number ratio by dividing all mole values by the smallest mole value (0.95 moles):
P:O = 0.95 moles : 4.42 moles
Simplifying by dividing by 0.95: P:O = 1 moles : 4.65 moles
Rounding to the nearest whole numbers: P:O ≈ 1:5
So, the empirical formula is P2O5.
Problem 4:
A compound contains 60% calcium, 13.3% sulfur, and 26.7% oxygen by mass. Determine its empirical formula.
Solution 4:
Follow the same steps as in Problem 1:
Convert the percentages to moles:
Calcium (Ca): 60% = 60 grams
Sulfur (S): 13.3% = 13.3 grams
Oxygen (O): 26.7% = 26.7 grams
Calculate the moles of each element:
Moles of Ca = 60 g / 40.08 g/mol (molar mass of Ca) ≈ 1.50 moles
Moles of S = 13.3 g / 32.07 g/mol (molar mass of S) ≈ 0.41 moles
Moles of O = 26.7 g / 16.00 g/mol (molar mass of O) ≈ 1.67 moles
Find the simplest whole number ratio by dividing all mole values by the smallest mole value (0.41 moles):
Ca:S:O = 1.50 moles : 0.41 moles : 1.67 moles
Simplifying by dividing by 0.41: Ca:S:O = 3.66 moles : 1 moles : 4.07 moles
Rounding to the nearest whole numbers: Ca:S:O ≈ 4:1:4
So, the empirical formula is CaSO4.
Problem 5:
A compound is found to contain 40% nitrogen and 60% oxygen by mass. Determine its empirical formula.
Solution 5:
Follow the same steps as in Problem 1:
Convert the percentages to moles:
Nitrogen (N): 40% = 40 grams
Oxygen (O): 60% = 60 grams
Calculate the moles of each element:
Moles of N = 40 g / 14.01 g/mol (molar mass of N) ≈ 2.85 moles
Moles of O = 60 g / 16.00 g/mol (molar mass of O) ≈ 3.75 moles
Find the simplest whole number ratio by dividing all mole values by the smallest mole value (2.85 moles):
N:O = 2.85 moles : 3.75 moles
Simplifying by dividing by 2.