The method: use moles to cross the equation

Use molar mass to convert grams to moles or moles to grams. Then use the coefficients in the balanced equation to compare different substances. Each conversion factor equals one in the relevant unit relationship, so orient it to cancel the unit you want to remove.

moles = mass (g) / molar mass (g/mol)

Coefficients give mole ratios. The small subscripts inside a formula tell you how many atoms belong to that formula unit. Keep these two jobs separate when finding a molar mass or building a reaction ratio.

The 2025 tables give atomic masses on page 1 and the mole–mass relationship on page 4. The balanced reaction and the material requirement come from the question’s cluster. Open the 2025 Reference Tables guide · Read the official periodic table

Try the official question

Official NYSED sample · Smartphone Chemistry · Question 3

Selected stimulus needed for this calculation. The full sample cluster is linked below.

A popular smartphone has a mass of 172 g and requires 10.32 g of aluminum to manufacture. Equation 1 represents a model of the overall chemical process used to obtain aluminum from the purified aluminum oxide.

Equation 1: 2Al₂O₃ + 3C → 4Al + 3CO₂

3. Construct a mathematical representation and calculate the number of grams of aluminum oxide required to produce the aluminum necessary to manufacture one smartphone. [1]

Mathematical Representation · g

Try a setup and an answer before reading the solution. NYSED sample booklet, page 2 · See the complete Smartphone Chemistry cluster

Worked reasoning · New Regents explanation

1. Identify the known mass and the requested substance

The question asks for aluminum oxide, Al₂O₃, needed to produce the aluminum in one phone. Use 10.32 g Al. The phone’s total mass of 172 g includes other materials and is not the aluminum mass.

2. Find the two molar masses in the 2025 table

Reference-table atomic masses and the molar masses used in this solution
SubstanceCalculationMolar mass
AlAtomic mass of Al26.98 g/mol
Al₂O₃2(26.98) + 3(16.00)101.96 g/mol

The periodic table gives Al as 26.98 and O as 16.00. One mole of Al₂O₃ contains two moles of Al atoms and three moles of O atoms, so add their mass contributions. The coefficient in front of Al₂O₃ in the reaction does not change its molar mass.

3. Convert the given aluminum mass to moles

10.32 g Al ×1 mol Al26.98 g Al≈ 0.3825 mol Al

The g Al units cancel. Keep the full calculator value for the next step, even though a rounded intermediate is displayed here.

4. Apply the mole ratio in the correct direction

The reaction shows 2 mol Al₂O₃ for every 4 mol Al. Aluminum is what we already know; aluminum oxide is what we need. Put Al on the bottom so its mole unit cancels.

n(Al₂O₃) = n(Al) ×2 mol Al₂O₃4 mol Al≈ 0.1913 mol Al₂O₃

This ratio comes from the coefficients in the whole balanced reaction. It is a ratio of moles, not a ratio of masses.

5. Convert aluminum oxide moles to grams

m(Al₂O₃) = n(Al₂O₃) ×101.96 g Al₂O₃1 mol Al₂O₃≈ 19.50 g Al₂O₃

The mol Al₂O₃ units cancel, leaving the requested grams of aluminum oxide. The unrounded value is 19.500133 g; rounding at the end gives 19.50 g.

6. Show the whole mathematical representation

10.32 g Al ×1 mol Al26.98 g Al×2 mol Al₂O₃4 mol Al×101.96 g Al₂O₃1 mol Al₂O₃≈ 19.50 g Al₂O₃

A sequence of separate calculations is also valid when it shows the same conversions and correct answer. As a reasonableness check, the required oxide has more mass than the aluminum alone because the oxide also contains oxygen. The mass ratio is not the coefficient ratio.

What earns the credit

This question asks for both a mathematical representation and a calculated answer. That differs from a prompt that asks only for a setup.

Official NYSED scoring guidance · Question 3

Allow 1 credit for 19.50 g. Both the mathematical representation and calculated answer must be correct to receive credit. Acceptable mathematical representations include, but are not limited to: 10.32 g Al× (1mole Al)/(26.98 g Al) × (2 moles Al₂O₃)/(4 moles Al) × (101.96 g Al₂O₃)/(1mole Al₂O₃) 10.32/26.98 = 0.3825 mol Al 0.3825× 2/4 = 0.1913 mol Al₂O₃ 0.1913 = x/101.96 Note: Significant figures do not need to be shown.

New Regents interpretation: Show a valid conversion path and the correct mass. The guide accepts equivalent representations and explicitly says significant figures do not need to be shown. Keep units in your work so you can check that each conversion goes the right way.

Read the original NYSED sample rating guide, page 1

Common mistakes to catch

  • Using the whole phone’s mass. The known amount is 10.32 g of aluminum, not 172 g of mixed phone materials.
  • Multiplying grams directly by the coefficient ratio. Convert Al to moles before comparing substances in the reaction.
  • Reversing the mole ratio. Put the substance you have in the denominator and the substance you want in the numerator. Here, mol Al must cancel.
  • Confusing coefficients with subscripts. Use Al₂O₃’s subscripts to calculate its molar mass; use the balanced reaction’s coefficients for the reaction mole ratio.
  • Writing only the final number. The official scoring guide requires the mathematical representation as well as the calculated answer.

Practice the skill in a complete cluster

Work the complete Smartphone Chemistry sample to connect this calculation with the phone’s glass, materials and electron-transfer questions. The full sample is free. You have seen Q3’s solution, so treat a repeat as supported practice.

Practice Smartphone Chemistry freeTry balloon stoichiometry

For a different mass-to-mass reaction problem, Balloon Arches includes a carbon-dioxide mass calculation. That full-exam workspace cluster requires membership; its public overview and the official exam PDF are available without an account.

All free Chemistry sample clusters · 2025 Reference Tables guide · Try the combined gas law lesson · Review another topic