Molar mass from percent composition, then a standard solution
SCH3U Grade 11 Chemistry · Quantities in Chemical Reactions
At the back of a greenhouse stockroom stands a jar of white crystals whose label has long since flaked away, and beside it on the bench sit the two reports that came back from the commercial lab. The elemental analysis puts the solid at 46.65% nitrogen by mass, and the infrared work confirms two nitrogen atoms per formula unit — the chemistry points to urea. To settle it and put the jar back to work, a technician zeroes an analytical balance, weighs a 12.00 g sample onto a creased square of weighing paper, and funnels it into a volumetric flask. Distilled water washes the crystals in, a gentle swirl dissolves them, and the flask is made up to the mark at 250.0 mL. Taking nitrogen as 14.01 g/mol, find the molar mass the analysis implies, the moles in the 12.00 g sample, the number of formula units that represents, and the concentration of the made-up solution.
Every number in this problem is editable — change any value below and the whole chain recalculates.
- pct_N = 46.65 % — Nitrogen by mass, from the analysis
- a = 2 atoms — Nitrogen atoms per formula unit
- M_N = 14.01 g/mol — Molar mass of nitrogen
- m = 12 g — Sample dissolved
- V = 250 mL — Volumetric flask made up to the mark
- (a)the molar mass the analysis implies
- (b)the moles in the 12.00 g sample
- (c)the number of formula units that represents
- (d)the concentration of the made-up solution
Run percent composition backwards. Two nitrogens weigh 28.02 g per mole of compound, and if that is 46.65% of the whole, the whole is fixed — this is how an analytical lab hands you a molar mass.
Carried onward at full precision, not this rounded figure.
The molar mass from step 1 carries in as M. Note what is being trusted here: an error in the reported percent nitrogen propagates into every number below it.
Carried onward at full precision, not this rounded figure.
Multiply by the Avogadro constant for the literal count of molecules — about 1.20 × 10²³. It changes nothing downstream; it is here because it is the only step that makes 12.00 g feel like a number of things.
Carried onward at full precision, not this rounded figure.
The sample is washed into a 250.0 mL volumetric flask and made up to the mark. Moles over volume of solution — not volume of water added — gives the concentration.
Carried onward at full precision, not this rounded figure.
Therefore the analysis pins the molar mass at 60.06 g/mol — urea, as the infrared suggested — the 12.00 g sample is 0.1998 mol, some 1.20 × 10²³ formula units, and made up to 250.0 mL it stands at 0.799 mol/L.
Why this order
Percent composition is usually taught in the forward direction, compound to percentages, because that is the easy direction. The useful direction is this one. A combustion analyser or a Kjeldahl digestion measures the mass fraction of one element; combine it with how many atoms of that element the formula holds and the molar mass drops out. That is the whole logic of empirical-formula determination, and it is why step 1 must come first — every later step needs an M, and there is no periodic table entry for an unlabelled jar.
Two things go wrong here reliably. The first is the subscript: it is 2 × 14.01, not 14.01, and forgetting the coefficient halves the molar mass without producing anything that looks obviously wrong. The second is the unit trap this whole shard is built around — molar mass is canonically kilograms per mole, so urea is 0.06006 kg/mol and not 60.06. Enter g/mol and the site converts; hand-enter 60.06 into an SI expression and you are out by a factor of a thousand. Step 4's volume is the flask, not the water: 12.00 g of urea dissolved and topped up to 250.0 mL is 0.799 mol/L, while 12.00 g stirred into 250.0 mL of water is a slightly larger volume and a slightly smaller concentration.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.