# How to Solve Chemistry Problems: A 6-Step Method

A repeatable 6-step method for solving chemistry problems, with a full stoichiometry example, a common wrong turn, a practice question and answer checks.

Published by BoredProf. 2026-11-04.

You’ve read the stoichiometry question three times. You know it involves grams, moles and a balanced equation, and you have no idea which to touch first. Here is the short answer: stop hunting for the right formula and run the same six steps every time. They turn a wall of text into small decisions you can actually make.

BoredProf publishes this article. It uses BoredProf as one example of a way to talk through your own course problems, and says so where it does.

The 6-step answer box

- Identify the question and the known quantities. Write what you are asked for (with its unit) and every number you are given (with its unit and substance).

- Draw or label the system. Sketch the reaction, the container or the particles, and label each quantity on the picture.

- Select the governing relationship. Pick the equation or concept that links what you know to what you want: a mole ratio, a gas law, a concentration formula.

- Track units. Write every unit, including the “of what” part, and set up conversions so they cancel.

- Calculate. Carry extra digits through the steps and round once, at the end, to the right significant figures.

- Check plausibility. Estimate the answer roughly, compare it with your result, and ask whether it makes chemical sense.

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## TL;DR

- Use the same six steps on every problem: question and knowns, sketch, relationship, units, calculate, plausibility check.

- Balanced-equation coefficients compare moles, never grams. Convert grams to moles first, every time.

- A ballpark estimate before or after calculating catches most unit slips and order-of-magnitude errors.

- After you solve a problem, explain why each step followed. That habit matters more than the number of problems you do.

- If you get stuck, talk the problem through step by step, with a TA, a help center, or a tool that explains out loud.

## Why a method beats hunting for formulas

Chemistry courses hand you dozens of problem types, and it is tempting to memorize one recipe per type. That breaks the moment a professor words a question differently.

A general method works because it asks the same questions regardless of topic. Many university courses teach exactly this. The Rutgers general chemistry course on LibreTexts uses [read, plan, solve, check](https://chem.libretexts.org/Courses/Rutgers_University/Chem_160%3A_General_Chemistry/02%3A_Matter_Measurement_and_Problem_Solving/2.07%3A_Solving_Problems_Involving_Equations): list the given information, identify the equations you need, confirm and convert units before calculating, then ask whether the answer is reasonable and complete. The Royal Society of Chemistry’s education site suggests a similar checklist called [GUESS](https://edu.rsc.org/ideas/success-with-guess-for-solving-equation-based-problems/4020659.article): Given, Units, Equation, Substitute, Solve.

The six steps above are the same idea with a sketch added, because many chemistry errors begin with a system you never pictured.

Stanford’s chemistry department adds a point worth taping to your desk. It advises that you [explain why and when each step is made](https://chemistry.stanford.edu/outreach/chemical-education/student-success/study-tips-chemistry), why certain information is useful or unnecessary, and why a given formula applies, rather than simply racking up problem counts. So in the example below, every step comes with its “because”.

## A full worked example: grams of ammonia from hydrogen

This is an original problem, built for practice.

Problem. Nitrogen and hydrogen react to form ammonia. If 10.0 g of H₂ reacts with excess N₂, how many grams of NH₃ can form?

### Step 1: Identify the question and the knowns

- Asked: mass of NH₃, in grams.

- Given: 10.0 g of H₂. N₂ is in excess, so H₂ limits the amount of product.

Why this step comes first: the unit you are asked for tells you where the chain of conversions must end. The phrase “excess N₂” tells you that you only need to follow H₂.

### Step 2: Draw or label the system

Write the unbalanced reaction, then balance it:

N₂ + 3 H₂ → 2 NH₃

Count atoms to confirm: 2 N and 6 H on each side. Label under each species what you know: H₂ is 10.0 g, NH₃ is the unknown.

Why: you can’t pick a mole ratio from an equation that isn’t balanced. The coefficients are the ratio.

### Step 3: Select the governing relationship

Two relationships do the work here:

- Molar mass converts between grams and moles of a single substance.

- The balanced equation’s coefficients give the mole ratio between different substances. Here that’s 2 mol NH₃ for every 3 mol H₂.

This is the core of reaction stoichiometry, covered in [OpenStax Chemistry 2e, section 4.3](https://openstax.org/books/chemistry-2e/pages/4-3-reaction-stoichiometry).

### Step 4: Track units

The route is: grams H₂ → moles H₂ → moles NH₃ → grams NH₃.

The University of Calgary’s chemistry textbook recommends writing out the “of what” part of every unit, so you never cancel units that look alike but describe different substances. “mol H₂” and “mol NH₃” do not cancel with each other.

The setup:

10.0 g H₂ × (1 mol H₂ / 2.016 g H₂) × (2 mol NH₃ / 3 mol H₂) × (17.031 g NH₃ / 1 mol NH₃)

Check the cancellation: g H₂ cancels, mol H₂ cancels, mol NH₃ cancels, and g NH₃ is left. The units land where Step 1 said they should.

### Step 5: Calculate

- Moles of H₂: 10.0 ÷ 2.016 = 4.960 mol

- Moles of NH₃: 4.960 × 2/3 = 3.307 mol

- Mass of NH₃: 3.307 × 17.031 = 56.32 g

The given mass has three significant figures, so the answer is 56.3 g NH₃.

Why you keep extra digits in the middle: Western Oregon University’s preparatory chemistry text says that for multiplication and division you [round to the fewest significant figures](https://cdn.wou.edu/chemistry/files/2017/03/CH150-Chapter-1-Final.pdf), and for multi-step problems you track precision without rounding intermediate values, rounding only the final answer. The molar masses and the 2/3 ratio come from exact counting or from the periodic table with more digits, so they don’t limit you. The 10.0 g does.

### Step 6: Check plausibility

Do a quick estimate with round numbers. In the balanced equation, 3 mol of H₂ is about 6 g, and it makes 2 mol of NH₃, about 34 g. We have 10 g of H₂, which is a bit under twice 6 g. So we expect something a bit under twice 34 g, say 55 to 60 g. Our 56.3 g fits.

There is a second sanity check: the product can’t weigh less than the hydrogen that went into it, since the nitrogen’s mass joins in. 56.3 g is well above 10.0 g. Good.

An academic paper on [using estimates in solving chemical problems](https://files.eric.ed.gov/fulltext/ED623175.pdf) makes the same point: chemical quantities have physical meaning, so estimating to the order of magnitude helps you recognize an unreasonable result.

## Why coefficients compare moles, not grams

This is the idea behind the most common stoichiometry mistake, so it deserves its own explanation.

A balanced equation is a statement about counting particles. N₂ + 3 H₂ → 2 NH₃ says: for every one N₂ molecule, three H₂ molecules react, and two NH₃ molecules form. A mole is just a very large, fixed count of particles. So the coefficients scale up cleanly to moles: 1 mol N₂ with 3 mol H₂ gives 2 mol NH₃.

Grams are different. Each molecule of a different substance has a different mass. One H₂ molecule is much lighter than one N₂ molecule. So 3 mol of H₂ is about 6 g, while 1 mol of N₂ is about 28 g. The coefficient 3 does not mean “3 g”, and it does not mean “three times the mass”. It means three times the count.

The molar mass is the bridge that turns a count into a mass. That’s why the route always goes through moles in the middle: grams of A to moles of A, then moles of A to moles of B using the coefficients, then moles of B to grams of B.

## A common wrong turn, and how to diagnose it

Here’s what many students do on the problem above:

10.0 g H₂ × (2 / 3) = 6.67 g NH₃

It looks tidy. It is wrong. How would you catch it?

Diagnosis 1: the units. Write them out. You have g H₂ times a ratio of “2 mol NH₃ per 3 mol H₂”. The grams of H₂ cannot cancel against moles of H₂. The leftover is a mess of mixed units, and the unit check fails before you ever reach a number.

Diagnosis 2: the plausibility check. The result, 6.67 g of NH₃, is less than the 10.0 g of hydrogen you started with. Ammonia contains all that hydrogen plus nitrogen, so its mass must be larger. The result is impossible.

The repair: go back to Step 4 and insert the molar-mass conversion on both sides of the mole ratio. The wrong turn comes from treating coefficients as masses, so the fix is to put “mol” next to every coefficient you use.

The Calgary advice about writing the full unit is what catches this. If you only wrote “g × 2/3”, nothing would look off.

Keep a short error log too. After two or three assignments, patterns show up: dropped units, skipped balancing, missed limiting reagents. Stanford also suggests [retrying every missed problem](https://chemistry.stanford.edu/outreach/chemical-education/student-success/study-tips-chemistry) before looking at the solution, to see whether the error was conceptual, mathematical, or stress.

## Quick check of understanding

Cover the answer and try this one.

Practice. Hydrogen burns in oxygen: 2 H₂ + O₂ → 2 H₂O. How many grams of water form from 4.00 g of O₂, with excess H₂?

Answer, with the reasoning:

- Asked: grams of H₂O. Given: 4.00 g O₂. Excess H₂ means O₂ limits.

- The equation is already balanced: 2 H₂ + O₂ → 2 H₂O.

- You need the molar mass of O₂ (32.00 g/mol), the ratio of 2 mol H₂O per 1 mol O₂, and the molar mass of H₂O (18.015 g/mol).

- Route: g O₂ → mol O₂ → mol H₂O → g H₂O.

- 4.00 g O₂ × (1 mol O₂ / 32.00 g) × (2 mol H₂O / 1 mol O₂) × (18.015 g / 1 mol H₂O) = 0.125 mol O₂, then 0.250 mol H₂O, then 4.504 g. With three significant figures: 4.50 g H₂O.

- Check: 32 g of O₂ makes 36 g of water (2 × 18), so 4 g makes 4.5 g. It matches. Also, water’s mass should exceed the O₂ mass, since it includes hydrogen. It does.

If you got 2.00 g by writing 4.00 × 1/2, look for the grams-as-coefficients mistake from the last section.

## Adapting the six steps to other problem types

The stoichiometry example is one pattern. The same steps work elsewhere, with a different Step 3.

- Solution concentration (molarity): the relationship is moles of solute divided by liters of solution. Step 4 usually involves converting milliliters to liters, which is where many order-of-magnitude errors start.

- Gas laws: the relationship is a gas law or the ideal gas equation. Step 4 means making sure temperature is in kelvin and that pressure and volume units match the gas constant you chose.

- Dimensional analysis alone: OpenStax describes the factor-label method as [units undergoing the same mathematical operations as their numbers](https://openstax.org/books/chemistry-2e/pages/1-6-mathematical-treatment-of-measurement-results), with each factor oriented so units cancel into the one you want.

- Conceptual questions: Steps 1 and 3 do the heavy lifting. Underline what the question asks you to do (calculate, explain, predict) and name the principle it points to.

A different kind of problem, assigning R and S at a stereocenter, has no calculation, but the same discipline applies: state what’s asked, label the system, apply the rule, check. You can see that one handled in [this step-by-step example on assigning R and S](https://boredprof.com/examples/assign-r-and-s).

## Is there an AI that solves chemistry problems?

Yes, many apps and sites now take a typed question or a photo of a problem and return a solution. Searching for “a Photomath for chemistry” turns up a long list of them.

What matters is what you do with the output. If you only copy the final number, you skip the exact skill your exam will test. A better use is to treat the answer as a draft to audit against the six steps:

- Did it identify the same question and knowns that you did?

- Is the equation balanced before any ratio is used?

- Do the units cancel, with “of what” labels, all the way through?

- Does the final number pass your own ballpark estimate?

- Are the significant figures consistent with the given data?

If any answer is no, you have found either a mistake in the tool’s work or a gap in yours. Both are worth knowing about. Tutorials that explain the reasoning also help more than ones that return only an answer: LibreTexts’ PASS Chem, for instance, offers a [guided solution that explains the chemistry rationale](https://chem.libretexts.org/Courses/Thompson_Rivers_University/PASS_Chemistry_Book_CHEM_1500/01%3A_Introduction/1.2%3A_How_to_use_PASS_Chem) behind each step, with hints about unit conversions, significant figures and common mistakes.

## Talking a problem through with BoredProf

Sometimes the hardest part is that the explanation you need is about your course problem, not a textbook one. BoredProf is an AI whiteboard tutor for exactly that. You upload your lecture slides, a PDF, a Word or PowerPoint file, or a photo of a homework problem. It explains the ideas out loud and draws the steps and figures on a whiteboard. You can use voice chat, turn on captions, ask follow-up questions, and save lessons to revisit. It also gives feedback on handwritten working.

If you’re a student who learns best by interrupting with “wait, why did the mole ratio go there?”, that back-and-forth is the point. For a guided look at the wider options, see our guide to [choosing an online chemistry tutor](/blog/online-chemistry-tutor/).

BoredProf covers math, physics, chemistry and biology. It is built for understanding and for checking explanations. It is not a replacement for your course, your textbook or a human tutor, and it does not promise grades or perfect accuracy, so keep using the six-step audit on anything it says.

The free plan lets you work with one lecture from one course, within stated usage caps. Pro is 29.99 CAD per month and Max is 59.99 CAD per month, and both expand access. Check the [live pricing page](https://boredprof.com) for current limits, prices and eligibility. The terms require users to be 13 or older.

Want to see how the whiteboard works before you upload anything? The public demo uses prewritten sample answers drawn on the real board, and the [worked-examples library](https://boredprof.com/examples) shows more problems solved step by step.

## Where to get help when you’re stuck

A method helps, and sometimes you still need a person. Many universities run free help rooms. The University of Nevada, Reno, for example, has a [Chemistry Help Center](https://www.unr.edu/chemistry/student-resources/chemistry-help-center) with drop-in help from chemistry graduate teaching assistants and no appointment needed. Hours change by semester, so check your own department’s page. Office hours, review sessions and tutoring centers at your school work the same way.

Whichever help you use, bring your own attempt. Show your six steps up to the point where they fall apart. “I got to step 4 and don’t know which conversion factor to use” is a question someone can answer in two minutes.

## FAQ

### How do I solve chemistry problems easily?

Follow the same six steps each time: identify the question and knowns, sketch the system, choose the relationship, track units, calculate, and check plausibility. Keep the steps on a sticky note until they are automatic. Then explain, in one sentence per step, why it follows from the last.

### Why can’t I use the equation’s coefficients as grams?

Coefficients count particles, and a mole is a fixed count of particles. Different substances have different masses per mole, so grams don’t scale with coefficients. Convert grams to moles with molar mass, apply the mole ratio, then convert back to grams.

### How do I know if my chemistry answer is reasonable?

Estimate with rounded numbers before or after you calculate, and see whether the answer should be bigger or smaller than a given value. Look for red flags like a product that weighs less than a reactant, or a result off by a factor of 10 or 1,000, which usually signals a missed unit conversion.

### Is there an app that solves chemistry problems?

Yes, several apps take typed questions or photos. Use them as a way to compare against your own working rather than as an answer source. Check that the equation is balanced, the units cancel and the number passes your estimate. BoredProf lets you discuss your own course problem with voice and a whiteboard, starting with a free plan.

## Try it on your own problem

Pick one problem from your current problem set, run the six steps, and then talk it through. You can [try the public demo](https://boredprof.com/#demo) or [start the limited free plan](https://app.boredprof.com/?start=1).

## Sources

- [2.7: Solving Problems Involving Equations (Rutgers, LibreTexts)](https://chem.libretexts.org/Courses/Rutgers_University/Chem_160%3A_General_Chemistry/02%3A_Matter_Measurement_and_Problem_Solving/2.07%3A_Solving_Problems_Involving_Equations)

- [Success with GUESS for solving equation-based problems (RSC Education)](https://edu.rsc.org/ideas/success-with-guess-for-solving-equation-based-problems/4020659.article)

- [Study Tips for Chemistry (Stanford Chemistry)](https://chemistry.stanford.edu/outreach/chemical-education/student-success/study-tips-chemistry)

- [Reaction Stoichiometry, OpenStax Chemistry 2e, section 4.3](https://openstax.org/books/chemistry-2e/pages/4-3-reaction-stoichiometry)

- [Preparatory Chemistry, Chapter 1 (Western Oregon University)](https://cdn.wou.edu/chemistry/files/2017/03/CH150-Chapter-1-Final.pdf)

- [The Use of Estimates in Solving Chemical Problems (ERIC)](https://files.eric.ed.gov/fulltext/ED623175.pdf)

- [1.6 Mathematical Treatment of Measurement Results (OpenStax Chemistry 2e)](https://openstax.org/books/chemistry-2e/pages/1-6-mathematical-treatment-of-measurement-results)

- [1.2: How to use PASS Chem (LibreTexts)](https://chem.libretexts.org/Courses/Thompson_Rivers_University/PASS_Chemistry_Book_CHEM_1500/01%3A_Introduction/1.2%3A_How_to_use_PASS_Chem)

- [Chemistry Help Center (University of Nevada, Reno)](https://www.unr.edu/chemistry/student-resources/chemistry-help-center)