Stoichiometry gram-to-gram problems aren't that bad once you stop overcomplicating them
The core task is simple: you're given a mass of one substance and asked to find the mass of another substance in the same reaction. The path runs through moles. That's it. You never convert grams directly to grams. If your teacher or worksheet ever implies you can skip the mole step, they're wrong or you're misreading the problem. Most teachers host these on Google Classroom, Canvas, or their school's LMS. Some use worksheets from textbooks like Zumdahl or Tro. A few post PDFs on Chemistryland or ChemTeam. If you need a quick answer key to check your work, search for the worksheet title plus "key" or "answers" rather than the topic name. "Combustion of propane stoichiometry worksheet answers" pulls up something useful. The generic search tends to return spam sites filled with ads. Step one: write and balance the equation. If it's already balanced on the worksheet, verify it yourself. Teachers sometimes drop unbalanced equations intentionally to catch students who skip that step. Step two: convert the given grams to moles using the molar mass. Step three: use the mole ratio from the balanced equation to switch from the given substance to the target substance. Step four: convert those moles back to grams using the target substance's molar mass.
I keep that as a single dimensional analysis line whenever I can. It looks like this in practice: given grams × (1 mol / molar mass) × (coefficient target / coefficient given) × (molar mass / 1 mol) = answer grams The units cancel cleanly if you set it up right. The mole units cancel. The gram units cancel. You're left with grams of the thing you're solving for.
A specific example with actual numbers
Take a standard worksheet problem: how many grams of water form when 25.0 grams of hydrogen gas react with excess oxygen? The balanced equation is 2H + O 2HO. Molar mass of H is 2.016 g/mol. Molar mass of HO is 18.015 g/mol. The mole ratio of HO to H is 2:2, which simplifies to 1:1 but I always use the balanced coefficients directly so you don't second-guess yourself later. 25.0 g H × (1 mol H / 2.016 g H) × (2 mol HO / 2 mol H) × (18.015 g HO / 1 mol HO) = 223.5 g HO
Get the Full Details
Significant figures: the given value has three, so the answer rounds to 224 grams. If the worksheet says 223.5, they're probably not tracking sig figs. That's common on intro worksheets. I don't recommend submitting answers without proper sig figs on a real exam though.
The edge case that always trips people up
Limiting reactant problems show up on almost every grams-to-grams worksheet, usually halfway through. The trick is that you're given masses for both reactants. You have to calculate the product yield from each one separately. The smaller yield is your actual answer. The other reactant is in excess. I ran into a problem recently where the worksheet gave 10.0 g of Al and 10.0 g of Cl for the reaction 2Al + 3Cl 2AlCl. My first pass gave me 39.8 g of AlCl from the aluminum and only 25.3 g from the chlorine. The answer was 25.3 g. The chlorine was limiting. The worksheet answer key had 39.8 g because the author made a mistake. I caught it by checking both calculations instead of just trusting the first number I got. Always check both reactants. The answer key will be wrong more often than you expect.
Hydrates and the hidden mass trap
Some worksheets slip in a hydrate like CuSO·5HO without warning. If you use the anhydrous molar mass instead of the full hydrate mass, your answer will be off by a large margin. CuSO is 159.61 g/mol. CuSO·5HO is 249.68 g/mol. That's a fifty-six percent difference. I've seen students lose points on this exact problem repeatedly because they treated the hydrate like the pure salt. Always check whether the formula includes water of crystallization before you grab a molar mass from the periodic table. Occasionally a worksheet adds a purity percentage like "the sample is 85% pure" or asks for a percent yield. These are easy to mishandle. Treat purity as a multiplier on your given mass before you convert to moles. If you start with 15.0 g of a sample that is 85% pure, your actual reacting mass is 12.75 g. Apply that adjustment before the mole conversion. Percent yield applies at the very end. Multiply your theoretical yield by the percent yield expressed as a decimal. Doing it in the wrong order will give you an answer that looks plausible but is numerically wrong. Gram-to-gram stoichiometry assumes complete reactions and pure substances. Real lab work is messier. Yields drop. Impurities exist. Side reactions happen. The worksheet method will overestimate your product mass every time if you take it straight into a lab setting. For academic purposes it's fine. For anything beyond an intro chem class, you'll need to account for equilibrium, side products, and actual measured yields. There's no shortcut around that. It's just more work.
If you're stuck on a specific problem from your worksheet, post the balanced equation and the numbers you were given. I can walk through the setup. Answer keys online are hit or miss, so checking your own work against the method above is usually faster than hunting for a key that matches your teacher's version.
