The short answer
Photograph the problem with Socratic Solve. The AI reads formulas, subscripts and reaction arrows, then explains the solution step by step: what's reacting, how to balance it, how to convert to moles, and how to get to the asked-for quantity with units carried through every line.
Unstick tonight's chemistry set
Download Socratic Solve free and scan the problem you're staring at.
The chemistry problems students scan most
- Balancing equations, with the element-by-element bookkeeping shown, so you see why the coefficients are 2, 3 and 1 instead of guessing.
- Mole conversions: grams ↔ moles ↔ particles, with molar masses worked out explicitly.
- Stoichiometry: from "how many grams of product" to limiting reagent and percent yield.
- Solutions and concentration: molarity, dilutions.
- Gas laws: PV = nRT and friends, with unit conversions handled.
- Conceptual questions: periodic trends, bonding, naming compounds; typed or photographed.
How to balance a chemical equation step by step
Balancing means the same number of each atom on both sides, because atoms aren't created or destroyed. You only ever change the coefficients (the big numbers in front), never the subscripts. Take the combustion of propane:
C3H8 + O2 → CO2 + H2O
- Start with the element that appears in only one compound on each side, and leave elements that appear alone (like O2) for last. Carbon: 3 on the left, so put a 3 in front of CO2.
- Balance hydrogen next. 8 H on the left, 2 per water molecule, so put a 4 in front of H2O.
- Balance oxygen last. The right side now has 3 × 2 + 4 × 1 = 10 O atoms. O2 carries 2 each, so you need 5 O2.
- Count every atom to confirm.
C3H8 + 5O2 → 3CO2 + 4H2O
| Element | Reactants | Products |
|---|---|---|
| C | 3 | 3 (in 3CO2) |
| H | 8 | 8 (in 4H2O) |
| O | 10 (in 5O2) | 10 (6 + 4) |
If you end up with a fraction (say 7/2 O2 when burning ethane), multiply every coefficient by 2 to clear it. The same count-and-adjust method balances 4Fe + 3O2 → 2Fe2O3: 4 iron and 6 oxygen on each side.
Stoichiometry worked example: grams to grams
How many grams of CO2 form when 22.0 g of propane burns completely? Every grams-to-grams problem uses the same three-step bridge: grams → moles → mole ratio → grams.
- Molar masses (C = 12.011, H = 1.008, O = 15.999 g/mol): C3H8 = 3(12.011) + 8(1.008) = 44.10 g/mol. CO2 = 12.011 + 2(15.999) = 44.01 g/mol.
- Grams to moles: 22.0 g ÷ 44.10 g/mol = 0.4989 mol C3H8.
- Mole ratio from the balanced equation: 1 C3H8 gives 3 CO2, so 0.4989 × 3 = 1.497 mol CO2.
- Moles to grams: 1.497 mol × 44.01 g/mol = 65.9 g CO2 (three significant figures, matching 22.0 g).
Write units on every line and let them cancel. If you're left with g2/mol or something equally strange, a conversion was flipped.
Limiting reagent: which reactant runs out first?
Now 22.0 g of propane reacts with only 64.0 g of O2. How much CO2 can form?
- Convert both to moles: propane 0.4989 mol (from above); O2 = 64.0 ÷ 32.00 = 2.000 mol.
- Compare with the ratio: burning all 0.4989 mol of propane needs 5 × 0.4989 = 2.49 mol O2. You only have 2.000 mol, so oxygen is the limiting reagent, even though there are more grams of it.
- Calculate product from the limiting reagent only: 2.000 mol O2 × (3 CO2 ÷ 5 O2) = 1.200 mol CO2 × 44.01 g/mol = 52.8 g CO2. That's the theoretical yield.
- Excess left over: the oxygen consumes 2.000 ÷ 5 = 0.400 mol propane (17.6 g), leaving about 4.4 g unreacted.
If the lab actually collects 45.0 g of CO2, the percent yield is 45.0 ÷ 52.8 × 100 = 85.2%.
Common chemistry mistakes to watch for
- Changing subscripts to balance. Turning H2O into H2O2 balances the oxygen but makes hydrogen peroxide, a different substance. Only coefficients change.
- Using a gram ratio instead of a mole ratio. The coefficients in a balanced equation compare moles (or molecules), never grams. Convert to moles first, every time.
- Picking the limiting reagent by mass. The reactant with fewer grams isn't necessarily the one that runs out; in the example above, oxygen had more grams and still ran out.
- Forgetting diatomic elements. H2, N2, O2, F2, Cl2, Br2 and I2 exist as pairs. Writing plain O in an equation throws off everything after it.
- Molar masses with parentheses. Ca(OH)2 has two O and two H: 40.078 + 2(15.999 + 1.008) = 74.09 g/mol.
- mL versus L. Molarity is moles per liter. 250 mL is 0.250 L.
- Significant figures. Round only at the end, and match the least precise measurement given. Rounding every intermediate step can shift your final digit.
Chemistry quick reference
| What you need | Formula or value |
|---|---|
| Moles from mass | n = mass ÷ molar mass |
| Particles from moles | particles = n × 6.022 × 1023 (Avogadro's number) |
| Molarity | M = moles of solute ÷ liters of solution |
| Dilution | M1V1 = M2V2 |
| Ideal gas law | PV = nRT, R = 0.08206 L·atm/(mol·K); temperature in kelvin (K = °C + 273.15) |
| Molar volume of a gas | 22.4 L/mol at 0 °C and 1 atm (check which STP your course uses) |
| Percent yield | actual ÷ theoretical × 100 |
Why step-by-step matters more in chemistry than anywhere
Stoichiometry is a chain: formula → balanced equation → mole ratio → conversion → answer. Get one link wrong and everything downstream is wrong, even with perfect arithmetic. When the app writes out each link, you can put your finger on the exact step where your attempt diverged (usually the mole ratio) and fix that one skill instead of "being bad at chemistry."
That's also why chemistry is the best subject for the check-before-you-submit routine: comparing chains line by line catches the single broken link in seconds.
Building real chemistry confidence
- Scan a problem and read the full solution once.
- Redo it on paper immediately, no peeking.
- Scan the next problem only after attempting it, then compare with the History tab's saved solution.
- Before the test, re-solve everything in your chemistry History cold.
Free to try: download Socratic Solve on the App Store. It includes free solve credits, so tonight's problem set costs nothing.