Chemistry

Chemistry Problem Solver with Steps

Balancing, stoichiometry, molarity, and gas laws with every conversion factor written out and the units cancelled on the page.

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First solution free

Nearly every calculation in an introductory chemistry course is the same journey with different scenery: get to moles, cross the balanced equation using the coefficient ratio, then convert to whatever the question asked for. Once that shape is visible, a limiting reagent problem stops looking different from a simple mass-to-mass one — it is the same route run twice with a comparison in the middle.

Working here is written as dimensional analysis, with each conversion shown as a fraction whose unwanted unit cancels. That is deliberate: the cancellation is what tells you the factor is the right way up, and an upside-down conversion factor is the most common error in the subject.

What this solver handles

Balancing by inspection and redox balancing by half-equations in acidic or basic solution.
Mole conversions between mass, particle count, and gas volume, with scientific notation handled properly around Avogadro's number.
Stoichiometry in both directions, including limiting reagent, excess remaining, theoretical yield, and percent yield.
Molarity, molality, dilution, and the mass of solute needed to prepare a stated solution.
Gas laws: Boyle, Charles, combined, ideal, and Dalton's law of partial pressures.
Empirical and molecular formulas from percentage composition or combustion data.
Acids and bases: pH, pOH, strong acid and base calculations, dilution of acids, and titration arithmetic.
Thermochemistry: heat capacity, latent heat, enthalpy of reaction, and Hess's law.
Word-style problems where the chemistry only starts after you unpack the paragraph and solve the resulting algebraic relationship.

Include the state symbols when they matter, since a reaction in aqueous solution and the same reaction between solids are different problems. Say whether conditions are standard, because the shortcut of 22.4 litres per mole only holds at STP. And if your data sheet quotes molar masses to a particular precision, use those figures — answers here are rounded to your least precise input, and periodic tables differ in the last decimal place.

Three problems, worked

Worked example

Balance C_{3} H_{8} + O_{2} \to CO_{2} + H_{2} O

1
Balance carbon first, since it appears in only one substance on each side. Three carbons on the left need three carbon dioxide molecules on the right.
$C_{3} H_{8} + O_{2} \to 3 CO_{2} + H_{2} O$
2
Hydrogen next, for the same reason. Eight hydrogens need four water molecules, because each water carries two.
$C_{3} H_{8} + O_{2} \to 3 CO_{2} + 4 H_{2} O$
3
Oxygen last, because it appears in two products. Count the right-hand side: six oxygens in the carbon dioxide and four in the water.
$3 (2) + 4 (1) = 10 oxygen atoms$
4
Ten oxygen atoms come from five diatomic oxygen molecules.
$C_{3} H_{8} + 5 O_{2} \to 3 CO_{2} + 4 H_{2} O$
5
Audit every element once more: 3 carbon and 3 carbon, 8 hydrogen and 8 hydrogen, 10 oxygen and 10 oxygen. Balanced.

Answer

C_{3} H_{8} + 5 O_{2} \to 3 CO_{2} + 4 H_{2} O

Worked example

Mass of NaCl in 250 mL of 0.40 M solution

1
Molarity is moles per litre, so the volume has to be in litres before it touches the formula. This single conversion is the most common failure point in solution chemistry.
$250 mL \times \frac{1 L}{1000 mL} = 0.250 L$
2
Multiply concentration by volume. Litres cancel, leaving moles.
$n = c V = (0.40 \frac{mol}{L}) (0.250 L) = 0.10 mol$
3
Find the molar mass by adding the atomic masses of sodium and chlorine.
$M = 22.99 + 35.45 = 58.44 \frac{g}{mol}$
4
Multiply moles by molar mass. Moles cancel and grams survive, which confirms the factor is the right way up.
$m = (0.10 mol) (58.44 \frac{g}{mol}) = 5.8 g$
5
Check for plausibility: a few grams of salt in a cup of water is about the saltiness of seawater, which is the right order of magnitude. Two significant figures, matching the 0.40.

Answer

5.8 g

Worked example

28.0 g N_{2} with 4.00 g H_{2} \to NH_{3} . Actual yield 18.0 g .

1
Write the balanced equation first — every ratio below depends on it.
$N_{2} + 3 H_{2} \to 2 NH_{3}$
2
Convert both reactant masses to moles using their molar masses.
$n (N_{2}) = \frac{28.0}{28.02} = 1.00 mol, n (H_{2}) = \frac{4.00}{2.02} = 1.98 mol$
3
Identify the limiting reagent by dividing each amount by its coefficient. Hydrogen gives the smaller quotient, so hydrogen runs out first even though there is more of it by mole count.
$\frac{1.00}{1} = 1.00 vs \frac{1.98}{3} = 0.66$
4
Use the limiting reagent and the coefficient ratio to find the moles of product. Two ammonia per three hydrogen.
$n (NH_{3}) = 1.98 \times \frac{2}{3} = 1.32 mol$
5
Convert to mass with the molar mass of ammonia, 14.01 plus three hydrogens.
$m = (1.32 mol) (17.03 \frac{g}{mol}) = 22.5 g$
6
Percent yield compares what was collected against that theoretical maximum. A yield above 100 per cent would mean the product is wet or impure, not that the chemistry over-performed.
$\frac{18.0}{22.5} \times 100 \approx 80%$

Answer

Theoretical 22.5 g, yield \approx 80%

Mass conservation checks a stoichiometry answer in a single line. The reaction consumed all 4.00 g of hydrogen and 0.66 mol of nitrogen, which is 18.5 g, for a total of 22.5 g — precisely the theoretical yield of ammonia. Atoms are neither created nor destroyed by a balanced equation, so if your product mass exceeds the mass of reactants actually consumed, a mole ratio has been inverted somewhere above.

Two further checks are worth running by reflex. A percent yield above 100 per cent means the product is wet, impure, or miscalculated, never that the reaction over-performed. And a mole count for a laboratory quantity almost always lands between about 0.001 and 10, so a result of 100 moles from a quarter of a litre of solution is wrong on sight — which is exactly the error below.

Finding the route through a calculation

The mole is the hub, and every calculation is a path into it and out again. What you are given decides the entry point; what is asked decides the exit. The tree below is that map with the questions asked in order.

Which route does this problem take?

Is the equation balanced yet?

Where students go wrong

Changing a subscript to balance an equation

Turning $H_{2} O$ into $H_{2} O_{2}$ balances the oxygen and replaces water with hydrogen peroxide. Only the coefficients in front of a formula may change, because they count molecules; subscripts define what the molecule is.

Using millilitres as litres

Molarity is moles per litre, so a volume in millilitres must be divided by 1000 before use. The error is easy to spot afterwards because the answer is off by a factor of exactly a thousand — a result in kilograms where grams were expected is almost always this.

Assuming the smaller mass is the limiting reagent

Four grams of hydrogen contains twice as many moles as 28 grams of nitrogen, because hydrogen is fourteen times lighter. Limiting is decided by moles divided by the stoichiometric coefficient, never by mass. Convert both, then compare.

Skipping the mole ratio

Going from grams of a reactant straight to grams of a product using only the two molar masses ignores the equation entirely. The coefficient ratio is the only step in the chain that is actually chemistry — the rest is arithmetic with units.

A thousandfold error, hiding in plain sight

One of these lines is wrong. Click it.

Three of those four are bookkeeping rather than chemistry, which is characteristic of the subject at this level: the concepts are rarely the bottleneck, the conversions are. Writing each conversion as a fraction and naming the unit that cancels makes an inverted factor almost impossible to miss, because the surviving units go wrong before the number does. The same discipline carries directly into physics problems, where units decide as many marks as the reasoning.

Formulas worth knowing cold

Chemistry reference

Tap any formula with a derivation to see where it comes from.

The mole

Solutions and gases

Acids, energy, and yield

Practice

Four to try

Answers are checked here — nothing is sent anywhere.

1
$Balance Fe + O_{2} \to Fe_{2} O_{3} . Coefficient of O_{2} ?$
2
$Moles in 36.0 g of water (M = 18.02 g/mol)$
3
$Molarity of 0.50 mol in 2.0 L$
4
$pH of 0.010 M HCl$

Frequently asked questions

Is the chemistry solver free?

Your first solution is free and needs no account. A free account then gives three solutions a day, and Gauth Plus removes the cap for $11.99 a month with a three-day free trial.

Does it show the steps or just the answer?

Each conversion factor is written as a fraction with the unit that cancels visible, so you can see grams becoming moles becoming moles of the other substance rather than watching a number appear.

Can I photograph a reaction from my notes?

Yes. Handwritten equations with subscripts photograph well, and state symbols are read correctly. Check that any subscript numbers are clear in the image, since a smudged 2 changes the substance.

Does it balance redox equations?

Yes, by the half-equation method in acidic or basic solution, with oxidation numbers assigned first and the electrons balanced before the two halves are combined.

Will it use the right significant figures?

Answers are rounded to the least precise measurement you gave. Molar masses are treated as exact enough not to limit the result, which matches how most courses mark this.

What topics does it cover?

Balancing, stoichiometry, limiting reagents and yield, molarity and dilution, gas laws, empirical and molecular formulas, acids and bases, and thermochemistry including Hess's law.

Keep going

Exponent and radical rulesScientific notation without the factor-of-ten slips.Turning words into equationsFor the paragraph that has to become a stoichiometry setup.Physics solverThe same unit discipline, applied to forces and energy.Equation solver with stepsFor the algebra a dilution or equilibrium setup leaves behind.