Exercises — Law of definite proportions (Proust)

What we do: find the fixed $A:B$ ratio from Sample 1, then reuse it. Step 1 — ratio from Sample 1: $\frac{A}{B}=\frac{4.0}{1.0}=4.0$. Why: the law promises this number is identical for every sample of this compound. Step 2 — apply to Sample 2: $A=4.0\times B=4.0\times 2.0=8.0$ g. Answer: $8.0$ g of $A$.

Step 1: the required fixed ratio is $O:H=8:1=8.0$. Step 2: the claimed ratio is $\frac{7.0}{1.0}=7.0\ne 8.0$. Why this matters: a pure compound cannot show a different ratio. So either it is not pure water, the sample is a mixture, or there is measurement error. Answer: Not consistent — $7.0\ne 8.0$.

Step 1 — set up the fixed ratio: $\frac{Mg}{O}=\frac{3}{2}$, so $O=\frac{2}{3}Mg$. Why: the ratio is a fixed number; rearrange it to solve for the unknown. Step 2 — substitute: $O=\frac{2}{3}\times 12.0=8.0$ g. Answer: $8.0$ g of oxygen.

Step 1 — oxygen needed: $O=\frac{Cu}{4.00}=\frac{10.0}{4.00}=2.50$ g. Why: rearrange $Cu/O=4.00$ to $O=Cu/4.00$. Step 2 — total compound mass (conservation of mass): $CuO=Cu+O=10.0+2.50=12.5$ g. Why: Law of Conservation of Mass — combined mass equals the sum of the parts. Answer: $12.5$ g of CuO, using $2.50$ g of oxygen.

Step 1 — total per unit recipe: for $8$ parts O and $1$ part H, total $=8+1=9$ parts. Step 2 — percentage: $\%O=\frac{8}{9}\times 100=88.9\%$ (to 1 d.p.). Why a percentage? A fixed mass ratio forces a fixed percentage by mass — that is exactly what percentage composition measures. Answer: $88.9\%$ oxygen (and $11.1\%$ hydrogen).

Step 1 — compute each $Fe:S$ ratio. Why: definite proportions demands one identical number.

• Sample 1: $7.00/4.00=1.75$
• Sample 2: $14.00/8.00=1.75$
• Sample 3: $7.00/6.00=1.167$ Step 2 — compare. Samples 1 and 2 share $1.75$; Sample 3 differs. Answer: Samples 1 and 2 are the same compound. Sample 3 is a different substance (or an impure/mixed sample) — its ratio breaks the law.

Step 1 — who is in excess? For $5.0$ g H, the fixed ratio demands $O=8\times 5.0=40.0$ g. We only have $32.0$ g O — so oxygen runs out first; hydrogen is in excess. Why: the fixed ratio sets the exact demand, and the smaller supply limits the reaction. Step 2 — hydrogen actually consumed: $32.0$ g O reacts with $H=\frac{32.0}{8}=4.0$ g H. Step 3 — water formed (conservation of mass): $32.0+4.0=36.0$ g water. Step 4 — leftover H: $5.0-4.0=1.0$ g of hydrogen remains as excess. Key insight: the leftover H does not enrich the water — every water molecule still obeys $8:1$. (See the red "leftover" block in the figure.) Answer: $36.0$ g water $+1.0$ g H left over.

(a) Each oxide alone: Definite proportions is about one compound. Any sample of oxide X will always show $C:O=12.0:16.0=0.75$, and any sample of oxide Y will always show $12.0:32.0=0.375$. Each has its own fixed ratio — so yes, each obeys the law individually. (b) The two together: Fix the carbon at $12.0$ g and compare the oxygen masses: $\frac{O_{\text{in }Y}}{O_{\text{in }X}}=\frac{32.0}{16.0}=\frac{2}{1}.$ The oxygen masses combining with a fixed carbon mass are in the small whole-number ratio $2:1$. That is the Law of Multiple Proportions. Answer: (a) Yes, each oxide has its own fixed ratio. (b) With C fixed, the O masses are in ratio $2:1$ — a multiple-proportions relationship, not a violation of definite proportions. (X is CO, Y is CO${}_2$.)

Step 1 — mass ratio to mole ratio. Divide each element's mass share by its atomic mass. Why: mass tells us "how heavy," but atom counting needs "how many," and dividing mass by atomic mass converts grams into a count proportional to atoms (The Mole Concept).

• Nitrogen: $\frac{14}{14}=1$
• Hydrogen: $\frac{3}{1}=3$ Step 2 — read the ratio: $N:H=1:3$. Answer: empirical formula ${\text{NH}}_3$ (ammonia).

Step 1 — mass ratio in terms of atomic masses. For formula $MO$ there is one atom of each, so $\frac{\text{mass of }M}{\text{mass of }O}=\frac{1\cdot m_M}{1\cdot 16.0}=\frac{40.0}{16.0}.$ Why: the parent's master formula $\frac{m_A}{m_B}=\frac{n_A{m}_A^{atom}}{n_B{m}_B^{atom}}$ with $n_M=n_O=1$. Step 2 — solve: $m_M=16.0\times \frac{40.0}{16.0}=40.0$. Answer: atomic mass of $M=40.0$ (calcium — the oxide is CaO).

Step 1 — total parts: $40+12+48=100$ parts. Step 2 — mass per element (each is its fraction of $100$ parts, scaled to $50.0$ g):

• Ca: $\frac{40}{100}\times 50.0=20.0$ g
• C: $\frac{12}{100}\times 50.0=6.0$ g
• O: $\frac{48}{100}\times 50.0=24.0$ g Step 3 — check (Law of Conservation of Mass): $20.0+6.0+24.0=50.0$ g. ✓ Answer: Ca $=20.0$ g, C $=6.0$ g, O $=24.0$ g. (This compound is CaCO${}_3$, calcium carbonate.)

Step 1 — who limits? For $10.0$ g H the ratio demands $O=8\times 10.0=80.0$ g. Only $10.0$ g O is present, so oxygen is the limiting element. Step 2 — hydrogen consumed: $H=\frac{O}{8}=\frac{10.0}{8}=1.25$ g reacts. Step 3 — water formed: $10.0+1.25=11.25$ g water (still $O:H=8:1$ internally). Step 4 — leftover H: $10.0-1.25=8.75$ g hydrogen unreacted. Exposing the error: the water is never $1:1$ — it is locked at $8:1$. The student confused the masses put into the beaker (a mixture, any ratio) with the masses inside the compound (fixed by law). Answer: $11.25$ g water forms, $8.75$ g H is left over; the water is $8:1$, not $1:1$.