Difficulty: Introductory | Prerequisites: basic arithmetic, periodic table familiarity, concept of atomic mass
This lab sits at the heart of general chemistry: connecting what you weigh on a balance to the atoms and molecules doing the reacting. If you can convert grams to moles and then find whole-number ratios, you can determine the simplest (empirical) formula of a compound from raw experimental data. The skill underpins stoichiometry, limiting reagents, and every quantitative problem you will meet later in the course. You should already be comfortable reading the periodic table for atomic masses and doing basic division.
To find a compound's empirical formula, convert each element's mass to moles using its molar mass, then divide every mole value by the smallest one to get whole-number subscripts. The lab demonstrates this with magnesium oxide (MgO) and extends it to other ionic compounds and copper oxide.
Mole (mol)
The SI unit for amount of substance, equal to 6.022 x 10²³ particles (Avogadro's number). Think of it as the chemist's counting unit: just as a "dozen" means 12, a "mole" means 6.022 x 10²³.
Molar mass (g/mol)
The mass of one mole of a substance, numerically equal to its atomic or molecular weight from the periodic table. In simple terms, it tells you how many grams you need to weigh out to have exactly one mole of that element or compound.
Empirical formula (simplest formula)
The formula showing the smallest whole-number ratio of atoms of each element in a compound. In simple terms, it is the most reduced version of a compound's formula. For example, the empirical formula of glucose (C₆H₁₂O₆) is CH₂O.
Subscript
The small number written to the lower right of an element symbol in a chemical formula, indicating the number of atoms of that element per formula unit. If no subscript is written, it is understood to be 1.
Ionic compound
A compound formed by the electrostatic attraction between oppositely charged ions (a metal cation and a non-metal anion). Think of it as a substance held together because positive and negative charges attract each other, like Mg²⁺ and O²⁻ forming MgO.
Determine the mass of each element present in the compound.
If you are given percentages, assume a 100 g sample so the percentages become grams directly.
If you have a compound mass and one element's mass, find the other by subtraction (mass of oxygen = mass of compound minus mass of metal).
Convert each mass to moles by dividing by the element's molar mass from the periodic table.
moles = mass (g) / molar mass (g/mol)
Divide every mole value by the smallest mole value in the set.
This gives the mole ratio for each element.
Round the ratios to the nearest whole number to get the subscripts.
If a ratio is close to 0.5, multiply all ratios by 2. If close to 0.33 or 0.67, multiply all by 3. The goal is whole numbers.
Write the empirical formula using those whole-number subscripts.
Magnesium ribbon (shiny, light grey) is burned in a crucible. The oxygen from air combines with the magnesium to form a white powder, magnesium oxide.
Mass of Mg = mass of (crucible + cover + Mg) minus mass of (crucible + cover) = 36.79 g minus 36.58 g = 0.21 g
Mass of compound (MgO) = 36.94 g minus 36.58 g = 0.36 g
Mass of O = mass of compound minus mass of Mg = 0.36 g minus 0.21 g = 0.15 g
Moles of Mg = 0.21 g / 24.305 g/mol = 0.009 mol
Moles of O = 0.15 g / 16.00 g/mol = 0.009 mol
Mole ratio: Mg = 0.009/0.009 = 1, O = 0.009/0.009 = 1
Empirical formula: MgO (a 1:1 ratio)
For ionic compounds, identify the charges on each ion and combine them so the total charge is zero.
Magnesium ion: Mg²⁺ (Group 2 metals lose two electrons)
Oxide ion: O²⁻ (oxygen gains two electrons)
One Mg²⁺ balances one O²⁻, so the formula is MgO.
2 Mg(s) + O₂(g) → 2 MgO(s)
Note that oxygen exists as a diatomic molecule (O₂), so you need two Mg atoms to balance the two oxygen atoms released.
Example (a): 0.200 mol Al and 0.600 mol Cl
Divide by the smallest (0.200): Al = 1, Cl = 3
Empirical formula: AlCl₃
Example (b): 0.080 mol Ba, 0.080 mol S, 0.320 mol O
Divide by the smallest (0.080): Ba = 1, S = 1, O = 4
Empirical formula: BaSO₄
2.50 g of copper reacts with oxygen to form 2.81 g of copper oxide.
Mass of O = 2.81 g minus 2.50 g = 0.31 g
Moles of Cu = 2.50 g / 63.55 g/mol = 0.039 mol
Moles of O = 0.31 g / 16.00 g/mol = 0.019 mol
Divide by smallest (0.019): Cu = 0.039/0.019 ≈ 2, O = 0.019/0.019 = 1
Empirical formula: Cu₂O
\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\text{mole ratio} = \frac{\text{moles of element}}{\text{smallest moles value}}\text{mass of element B} = \text{mass of compound} - \text{mass of element A}Empirical formula determination is the basis of analytical chemistry: pharmaceutical companies use it to confirm the identity and purity of drug compounds, and forensic labs use it to identify unknown substances. Industrial processes rely on the same mass-to-mole conversions to calculate how much raw material is needed for a desired product yield.
Students often confuse the empirical formula with the molecular formula. The empirical formula is the simplest ratio; the molecular formula may be a whole-number multiple of it (e.g., CH₂O vs C₆H₁₂O₆). This lab only asks for the empirical formula.
Students sometimes divide by the largest mole value instead of the smallest. Always divide by the smallest to get ratios of 1 or above.
Forgetting that oxygen is diatomic (O₂) when writing balanced equations. The oxygen you breathe and that reacts in combustion is O₂, not single O atoms.
Rounding mole ratios too aggressively. A ratio of 1.5 is not "close enough" to 2. It means you should multiply all ratios by 2 (giving 3:2, for instance). Only round when the value is within about 0.1 of a whole number.
⚠️ You will almost certainly be asked to determine an empirical formula from mass data on the exam. Practise the three-step workflow (mass → moles → divide by smallest) until it is automatic.
⚠️ Balancing combustion and synthesis equations is a recurring exam topic. Remember: elements that exist as diatomic molecules (H₂, O₂, N₂, F₂, Cl₂, Br₂, I₂) must appear that way in balanced equations.
⚠️ Know the charges on common ions (Group 1 = +1, Group 2 = +2, oxygen = −2) so you can write correct ionic formulas without a reference sheet.
True or false: The empirical formula of a compound always shows the actual number of atoms in one molecule. (False, that is the molecular formula.)
Fill in the blank: To convert grams to moles, divide the mass by the element's ________. (molar mass)
True or false: When finding the mole ratio, you divide each mole value by the largest mole value. (False, divide by the smallest.)
Fill in the blank: Magnesium has a charge of ________ when it forms an ion. (2+ or Mg²⁺)
True or false: Oxygen gas in a balanced equation is written as O rather than O₂. (False, oxygen is diatomic: O₂.)
Q: A compound contains 0.050 mol of iron (Fe) and 0.075 mol of oxygen (O). What is its empirical formula?
A: Divide by the smallest (0.050): Fe = 1, O = 1.5. Multiply both by 2: Fe = 2, O = 3. Empirical formula: Fe₂O₃.
Q: 3.00 g of carbon reacts with oxygen to form 11.00 g of a compound. What is the empirical formula?
A: Mass of O = 11.00 minus 3.00 = 8.00 g. Moles of C = 3.00/12.01 = 0.250. Moles of O = 8.00/16.00 = 0.500. Ratio: C = 1, O = 2. Empirical formula: CO₂.
Q: Write the balanced equation for aluminium reacting with oxygen gas to form aluminium oxide (Al₂O₃).
A: 4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s).
Q: A student obtains a mole ratio of Cu:O = 1.98:1. What empirical formula should they report?
A: 1.98 is close enough to 2 (within rounding tolerance), so the empirical formula is Cu₂O.
This material connects directly to stoichiometry (using mole ratios from balanced equations to predict how much product forms) and limiting reagents (identifying which reactant runs out first). It also lays groundwork for percent composition, which reverses the process: given a formula, calculate what fraction of the compound's mass each element contributes.
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