Reacting Masses Gases And Yields

Finding moles of each element $${n}_{C}=\frac{40.0}{12.0}=3.33$$ $${n}_{H}=\frac{6.7}{1.0}=6.70$$ $${n}_{O}=\frac{53.3}{16.0}=3.33$$ Ratio C:H:O = 1:2:1, so the empirical formula is CH₂O. Finding the multiplier $${M}_{r}(\text{empirical})=30$$ $$n=\frac{180}{30}=6$$ The molecular formula is C₆H₁₂O₆.

Empirical formula gives the simplest whole-number ratio of atoms in a compound; molecular formula gives the actual number of atoms of each element in one molecule.

$$n(\text{Mg})=\frac{4.86}{24.3}=0.200\text{ mol}$$ $$n(\text{MgO})=0.200\text{ mol (1:1 ratio)}$$ $$m(\text{MgO})=0.200\times 40.3=8.06\text{ g}$$

$$n(\text{Na})=\frac{4.60}{23.0}=0.200\text{ mol}$$ $$n(\text{NaCl})=0.200{\text{ mol (1:1 ratio from 2Na + Cl}}_{2}\to \text{2NaCl)}$$ $$m(\text{NaCl})=0.200\times 58.5=11.7\text{ g}$$

$$n({\text{CaCO}}_{3})=\frac{5.00}{100.1}=0.0500\text{ mol}$$ $$n({\text{CO}}_{2})=0.0500\text{ mol}$$ $$V({\text{CO}}_{2})=0.0500\times 24.0=1.20{\text{ dm}}^{3}$$

$$n({\text{H}}_{2})=\frac{600}{24 000}=0.0250\text{ mol}$$ $$n(\text{Na})=0.0500\text{ mol (2:1 ratio)}$$ $$m(\text{Na})=0.0500\times 23.0=1.15\text{ g}$$

Converting to SI units: p = 101 000 Pa, V = 4.80 × 10⁻⁴ m³, T = 298 K. $$n=\frac{pV}{RT}=\frac{101 000\times 4.80\times {10}^{-4}}{8.31\times 298}$$ $$n=0.0196\text{ mol}$$

Converting to SI units: p = 100 000 Pa, V = 9.50 × 10⁻⁵ m³, T = 353 K. $$n=\frac{100 000\times 9.50\times {10}^{-5}}{8.31\times 353}=3.24\times {10}^{-3}\text{ mol}$$ $${M}_{r}=\frac{0.250}{3.24\times {10}^{-3}}$$ $${M}_{r}=77.2$$

$${M}_{r}(4{\text{H}}_{2})=8.0$$ $$\text{total }{M}_{r}(\text{products})=44.0+8.0=52.0$$ $$\%\text{ atom economy}=\frac{8.0}{52.0}\times 100=15.4\%$$

$$n(A)=\frac{10.0}{100}=0.100\text{ mol}$$ $$n(B{)}_{\text{theoretical}}=0.100\text{ mol}$$ $$m(B{)}_{\text{theoretical}}=0.100\times 80=8.00\text{ g}$$ $$\%\text{ yield}=\frac{6.40}{8.00}\times 100=80.0\%$$

$$m(\text{O})=4.03-2.43=1.60\text{ g}$$ $$n(\text{Mg})=\frac{2.43}{24.3}=0.100\text{ mol}$$ $$n(\text{O})=\frac{1.60}{16.0}=0.100\text{ mol}$$ Ratio Mg:O = 1:1, so the empirical formula is MgO; the equation 2Mg + O₂ → 2MgO is confirmed.

Some magnesium oxide may escape as white smoke; this lowers the apparent mass of oxide so the calculated O ratio is too low. Some unreacted magnesium may remain inside the oxide; this means less oxygen has reacted so the calculated O ratio is too low.

IV = mass of magnesium; DV = volume of hydrogen collected; CV = temperature/concentration of acid/excess acid present (any one).

Hydrogen is slightly soluble in water so some dissolves before reaching the cylinder; some gas may escape before the bung is fitted; both effects decrease measured V which lowers the calculated molar volume.

$$n(\text{Mg})=\frac{0.0729}{24.3}=3.00\times {10}^{-3}\text{ mol}$$ $$n({\text{H}}_{2})=3.00\times {10}^{-3}\text{ mol}$$ $${V}_{m}=\frac{72.0}{3.00\times {10}^{-3}}=24 000{\text{ cm}}^{3}{\text{ mol}}^{-1}$$ $${V}_{m}=24.0{\text{ dm}}^{3}{\text{ mol}}^{-1}$$ This matches the accepted RTP value, indicating no significant gas loss.

$${\text{Ba}}^{2+}(\text{aq})+{\text{SO}}_{4}^{2-}(\text{aq})\to {\text{BaSO}}_{4}(\text{s})$$ State symbols correct; observation: white precipitate forms.

$$2{\text{H}}^{+}(\text{aq})+{\text{MgCO}}_{3}(\text{s})\to {\text{Mg}}^{2+}(\text{aq})+{\text{H}}_{2}\text{O}(\text{l})+{\text{CO}}_{2}(\text{g})$$ State symbols correct; observations: effervescence/bubbles; solid dissolves/disappears.

The ionic equation is: $${\text{Cl}}_{2}(\text{aq})+2{\text{I}}^{-}(\text{aq})\to 2{\text{Cl}}^{-}(\text{aq})+{\text{I}}_{2}(\text{aq})$$ Chlorine is more reactive than iodine, so it displaces iodide ions; the brown colour is due to formation of aqueous iodine. Calculating the percentage yield: $$n(\text{Mg})=\frac{5.95}{24.3}=0.245\text{ mol}$$ $$n({\text{H}}_{2}{)}_{\text{theoretical}}=0.245\text{ mol}$$ $${V}_{\text{theoretical}}=0.245\times 24.0=5.88{\text{ dm}}^{3}$$ $$\%\text{ yield}=\frac{5.85}{5.88}\times 100=99.5\%$$ Some hydrogen dissolves in water during collection, or a small mass of Mg may remain unreacted on surface oxide.