Keep mass, amount and concentration separate
Mass describes the sample in units such as grams. Amount of substance in moles counts entities through the mole unit. Molar concentration relates an amount of solute to final solution volume. These quantities connect, but they are not interchangeable inputs.
For a sample with mass m and molar mass M, n = m/M. Using a stated molar mass of 58.5 g/mol, 11.7 g corresponds to 11.7/58.5 = 0.200 mol. Keep the substance and the units with the calculation so the amount is attached to the correct species.
Calculate molarity from final solution volume
Molar concentration c = n/V when n is in moles and V is the final solution volume in liters. The worked example uses 0.50 mol in 2.0 L, giving 0.25 mol/L. If 0.060 mol is in 300 mL of solution, convert 300 mL to 0.300 L, then calculate 0.060/0.300 = 0.20 mol/L.
Do not replace final solution volume with the amount of solvent added unless the problem establishes that relationship. At fixed solute amount, doubling the final volume halves the molarity. That prediction is a useful scale check alongside the arithmetic.
Balance with coefficients, preserving formulas
Consider the prepared equation N₂ + H₂ → NH₃. Place 2 before NH₃ to give two nitrogen atoms on each side, then 3 before H₂ to give six hydrogen atoms on each side. The balanced equation is N₂ + 3H₂ → 2NH₃. Count each element to check the result.
Changing NH₃ to NH₂ would change the represented substance. Balance by changing the coefficients, with subscripts kept intact. A coefficient multiplies the entire formula: 2NH₃ represents two nitrogen atoms and six hydrogen atoms in a molecular count.
Use the balanced equation as a mole relationship
For N₂ + 3H₂ → 2NH₃, one mole of N₂ reacts with three moles of H₂ and corresponds to two moles of NH₃. If 0.40 mol of N₂ reacts completely with hydrogen in excess, the theoretical amount of NH₃ is 0.40 × 2 = 0.80 mol. This is a mole ratio, not a mass ratio.
A theoretical calculation assumes the conditions specified by the problem, including which reactant limits the reaction. An actual yield can be lower. When converting a product amount to mass, use that product’s molar mass after applying the mole ratio.
Identify a limiting reactant by the same reaction
Suppose 0.40 mol N₂ and 0.90 mol H₂ are available for the same equation. The nitrogen would need 1.20 mol H₂ to react fully, but only 0.90 mol is available. Hydrogen limits the reaction. It can produce 0.90 × (2/3) = 0.60 mol NH₃ theoretically.
Alternatively, calculate the possible product amount from each reactant and choose the smaller result. Comparing the starting mole amounts directly is insufficient because the coefficients differ. Here, 0.90 mol H₂ consumes 0.30 mol N₂, leaving 0.10 mol N₂ in this idealized calculation.
Report what the numerical answer describes
A final answer should name the substance or solution, the quantity and the unit. Distinguish mol from mol/L and grams from moles. Follow the significant-figure and rounding instructions in the assignment, carrying enough digits through intermediate calculations.
If your result seems wrong, first check the formula, the final volume, the balanced coefficients and the conversion path. Ask Lirno about the earliest unclear relationship. These examples support written homework; practical experiments use your class’s approved procedures and supervision.
Go from a mass to a product amount in separate steps
For the prepared reaction 2Mg + O₂ → 2MgO, suppose 4.8 g of magnesium reacts completely with excess oxygen. Use the supplied rounded molar masses Mg = 24 g/mol and MgO = 40 g/mol. First convert magnesium mass to amount: 4.8/24 = 0.20 mol. The balanced equation has a 2:2 ratio, so this gives 0.20 mol MgO.
Convert the product amount back to mass: 0.20 × 40 = 8.0 g MgO. The product weighs more than the magnesium alone because oxygen contributes mass. With these rounded values, 3.2 g of oxygen is incorporated, and 4.8 + 3.2 = 8.0 g. Use the atomic masses supplied in your assignment when they differ; do not mix rounded and more precise values halfway through.
