Empirical formula practice problems become easier when you keep three things separate: mass, moles, and the final atom ratio. Convert each element’s mass to moles, divide every mole amount by the smallest, and check whether the entire ratio needs a whole-number multiplier. Don’t round a clear 1.5 to 2.
This practice set includes eight original problems, worked answers, and a way to diagnose the step that went wrong. It covers mass data, percentage composition, fractional ratios, and the extra information needed for a molecular formula. Use the questions as a short worksheet before opening the answers.
What an empirical formula tells you
An empirical formula gives the simplest whole-number ratio of atoms. A molecular formula gives the actual atom counts in a molecule. For example, reducing C4H8 to CH2 preserves the carbon-to-hydrogen ratio, but it doesn’t describe the same number of atoms. Some molecular formulas already have irreducible subscripts. See OpenStax’s explanation of chemical formulas for the distinction.
That distinction matters when you write your final answer. If a question asks only for an empirical formula, a larger equivalent ratio is unfinished. If it asks for a molecular formula, a simplest ratio alone may be insufficient. Underline the requested output before calculating.
The method to put beside your worksheet
- Identify the mass of each element. If all composition percentages are given, use an imaginary 100 g sample.
- Calculate moles of each element by dividing grams by that element’s atomic molar mass.
- Divide every mole amount by the smallest of those amounts.
- If necessary, multiply every part of the ratio by the same small integer.
- Write subscripts and check that they have no common factor greater than one.
This mass-to-mole procedure is described in OpenStax’s empirical and molecular formula section. The worked exercises below use deliberately selected teaching data. They are not experimental measurements or copied homework answers.
For this set, use C = 12.01, H = 1.008, O = 16.00, N = 14.01, Mg = 24.31, and Al = 26.98 g/mol. Keep extra calculator digits during conversion. Round displayed results when needed, while retaining enough precision to recognize a ratio.
Eight empirical formula practice problems
1. Start with two measured masses
A sample contains 4.862 g magnesium and 3.200 g oxygen. Find its empirical formula. Write the mole amount under each mass before comparing the elements.
2. Work from percentage composition
A compound contains 52.14% carbon, 13.13% hydrogen, and 34.73% oxygen by mass. Determine its empirical formula. Your first line should state your assumed sample mass.
3. Handle a ratio ending in 1.5
A compound contains 5.396 g aluminum and 4.800 g oxygen. Find its empirical formula. Explain why rounding the oxygen ratio to 2 would change the answer.
4. Recognize a thirds ratio
A compound contains 7.206 g carbon and 0.8064 g hydrogen. Determine the empirical formula. Your normalized ratio will include approximately 1.333.
5. Find a missing element’s mass first
A 7.407 g sample contains only carbon, hydrogen, and oxygen. Analysis gives 3.603 g carbon and 0.6048 g hydrogen. Find the oxygen mass, then the empirical formula.
6. Move from empirical to molecular formula
A molecular compound has empirical formula CH2O and molar mass approximately 90.08 g/mol. Calculate its molecular formula. Show the mass of one empirical-formula unit on a molar basis.
7. Check whether a ratio is already simplest
Reduce each molecular formula to its empirical formula: C6H14, C5H12, and N2O4. A subscript of 1 is implicit when no number is printed.
8. Decide when the data cannot support an answer
A worksheet gives CH2O as an empirical formula and 75.00 g/mol as the compound’s molar mass. It gives no uncertainty range. Can you confidently assign a molecular formula? Explain the problem before trying to force a whole number.
Worked answers and mistake checks
1. Magnesium and oxygen: MgO
Magnesium: 4.862 / 24.31 = 0.2000 mol. Oxygen: 3.200 / 16.00 = 0.2000 mol. Dividing both by 0.2000 gives 1:1, so the empirical formula is MgO.
The masses weren’t equal, but the mole amounts were. That’s the point of the conversion. A formula records relative atom counts, so using the original gram ratio would answer a different question. On your worksheet, circle the two mole amounts to make that change of quantity visible.
2. Percentage composition: C2H6O
Assume 100 g. The element masses are 52.14 g C, 13.13 g H, and 34.73 g O. Their mole amounts are approximately 4.341, 13.026, and 2.171. Dividing by the smallest gives approximately 2.000:6.001:1.
The empirical formula is C2H6O. Here the small deviations from integers reflect the rounded percentage inputs. Notice that the oxygen subscript is 1. The subscripts 2 and 6 sharing a factor doesn’t let you reduce the formula, because every subscript must share that factor.
3. Aluminum and oxygen: Al2O3
Aluminum: 5.396 / 26.98 = 0.2000 mol. Oxygen: 4.800 / 16.00 = 0.3000 mol. Dividing by 0.2000 produces 1:1.5. Multiply both numbers by 2 to get 2:3.
The answer is Al2O3. Rounding 1.5 upward would produce 1:2, a different proportion. Multiplying only the oxygen number would produce 1:3, also different. Write the multiplier across the whole ratio, like this: (1:1.5) × 2 = 2:3.
4. Carbon and hydrogen: C3H4
Carbon: 7.206 / 12.01 = 0.6000 mol. Hydrogen: 0.8064 / 1.008 = 0.8000 mol. Dividing by 0.6000 gives 1:1.3333…, equivalent to 1:4/3. Multiplying both terms by 3 gives 3:4.
The empirical formula is C3H4. This exercise tests the arithmetic of composition data; the ratio alone does not establish a substance’s structure. Don’t choose a familiar formula first and try to make the numbers fit it. Check the calculation and the assumptions separately.
5. Oxygen by subtraction: C3H6O2
Oxygen mass = 7.407 – 3.603 – 0.6048 = 3.1992 g. Keeping that unrounded intermediate value gives approximately 0.3000 mol C, 0.6000 mol H, and 0.19995 mol O. Divide each by 0.19995: approximately 1.5004:3.0008:1.
Multiply all three terms by 2 to obtain approximately 3:6:2. The empirical formula is C3H6O2. Subtraction was valid because the prompt explicitly said the sample contained only those three elements. If another element were present, assigning all leftover mass to oxygen would be unjustified.
6. Molecular formula: C3H6O3
The empirical molar mass is 12.01 + 2(1.008) + 16.00 = 30.026 g/mol. The multiplier is 90.08 / 30.026, approximately 3. Multiply every subscript in CH2O by 3 to get C3H6O3.
Check backward: 3(30.026) = 90.078 g/mol, consistent with the given value. Keep the empirical answer and molecular answer on separate labeled lines. That makes it harder to stop at CH2O when the question asks for the larger formula.
7. Reduction answers
C6H14 reduces by 2 to C3H7. C5H12 stays unchanged because 5 and 12 share no whole-number factor greater than one. N2O4 reduces by 2 to NO2.
A useful self-check is to multiply the reduced subscripts by your chosen divisor. You should recover every original subscript. If one element doesn’t return to its original count, you didn’t apply one common divisor to the entire formula.
8. Inconsistent molecular-formula information
The proposed multiplier is 75.00 / 30.026, approximately 2.498. That’s not close to an integer at the stated precision. Neither multiplying by 2 nor multiplying by 3 matches the supplied molar mass.
Report that the data need checking. Possible places to inspect include a copied molar mass, the empirical formula, or missing uncertainty information. You cannot identify which input is wrong from this prompt alone. A neat-looking final formula doesn’t repair inconsistent premises.
Turn wrong answers into useful chemistry notes
For each missed problem, record the first incorrect line, the corrected line, and one sentence explaining the change. For example: “I divided atomic molar mass by grams. I need grams divided by grams per mole so the result is moles.” This gives a study partner something more useful than an answer key.
If premature rounding caused the error, try the significant figures practice problems. For related course material, explore chemistry notes or browse study documents. Check another student’s atomic-mass assumptions before comparing decimal answers.
Upload your own notes to help other students. Include your original questions, labeled calculations, corrections, and source references. Share work you have permission to share, and keep restricted assessments and classmates’ personal information out of the file.
Frequently asked questions
Can I always assume a 100 g sample?
Use that convenient basis when the problem gives percentage composition. When it gives actual masses, use those masses directly. You don’t need to convert them to percentages first.
How do I choose the ratio multiplier?
In these exercises, halves need a factor of 2 and thirds need a factor of 3. Apply the factor to every element. Inspect the original precision before treating an arbitrary decimal as a simple fraction.
Where can I find another worked route?
Purdue’s stoichiometry practice index separates empirical formula problems by mass and percentage data, and also offers molecular formula examples. Pick the route that matches the information your homework gives you.