Chemistry Final Problem Set Cheat Sheet

↑
- 2023-06-12 1

Notes

Finding AMU
- C12 = 98.9
- C14 = 1.1
- $(12 * 98.9) + (14 * 0.11)$
- $12.022 a m μ$
Empirical -> Molecular
- $\frac{M a s s M o l e c u l a r}{M a s s E m p i r i c a l} = R a t i o$
- Find mols of each compound for hydrates or each atom. Divide by lowest.
- Then find empirical
Combustion Analysis
- $0.539 g$ of $C$ and $H_{2}$ >> $1.64 g$ of $C O_{2}$ and $0.807 g$ of $H_{2} O$
- $\frac{M o l a r M a s s H_{2}}{M o l a r M a s s H_{2} O} * 0.807 g H_{2} O = 0.0905 g H_{2}$
- $\frac{2.02 g H_{2}}{18.02 g H_{2} O} * 0.807 g H_{2} O = 0.0905 g H_{2}$
- $\frac{12.01 g C}{44.01 g C O_{2}} * 1.64 g C O_{2} = 0.4475 g C$
- //This is stoichiometry, times the mass to find mass of product.
- $\frac{0.0905 g}{1.01 g / m o l} = 0.0896 m o l$
- //The limiting reactant was $H_{2}$ so we used that to find the mols of $H_{2}$
Variables
- $N_{A} = 6.022 * 10^{23}$ //Avogadro's Number
- $n = M o l e s$
- $N_{A} = \frac{# o f P a r t i c l e s}{n}$
- $M = M o l a r M a s s$
- $m = G r a m s$
Percent Yield
- $P e r c e n t Y i e l d = \frac{A c t u a l Y i e l d}{T h e o r e t i c a l Y i e l d}$
Statements
- AMU is equivalent to g/mol
- Limiting reactant is x
Mass Percent
- $M a s s P e r c e n t Z n = \frac{M a s s Z n}{M a s s C o m p o u n d}$
Abundance
- $(A b u n d a n c e * m a s s 1) + (A b u n d a n c e * m a s s 2)$
Gas Law
- $P V = n R T$
- $R = 8.314$
- $P = P r e s s u r e$
- $V = V o l u m e$
- $n = M o l e s$
- $T = T e m p e r a t u r e$
- $\frac{P_{i} V_{i}}{T_{i}} = \frac{P_{F} V_{F}}{T_{F}}$
- $\frac{P_{i}}{T_{i}} = \frac{P_{F}}{T_{F}}$
- $\frac{V_{i}}{T_{i}} = \frac{V_{F}}{T_{F}}$
- $P_{i} V_{i} = P_{F} V_{F}$
- STP
  - 101.3kPa
  - 0c
  - 273K
- SATP
  - 100kPa
  - 25c
  - 298K
Converting Pressure
- $1 a t m = 760 t o r r = 760 m m H g = 101.3 k P a$
- $2 a t m * \frac{760 t o r r}{1 a t m} = 1520 t o r r$
Molar Volume
- $M o l a r V o l u m e = \frac{V}{n}$
$C = \frac{n}{V}$
- Concentration = number of moles/volume
Concentrations
- $C = \frac{n}{V}$
  - Concentration
  - Number of Moles
  - Volume
- $C = \frac{n}{M}$
  - Concentration
  - Number of Moles
  - Molarity = mol/L
- $C = \frac{V_{s o l u t e}}{V_{s o l v e n t}} = \frac{M a s s_{s o l u t e}}{V_{s o l v e n t}} = \frac{M a s s_{s o l u t e}}{M a s s_{s o l v e n t}}$
Dilution
- $(C_{1}) (V_{1}) = (C_{2}) (V_{2})$
- $p p m = \frac{s o l u t e}{s o l v e n t} * 10^{6}$
- $p p b = \frac{s o l u t e}{s o l v e n t} * 10^{9}$
- $p p t = \frac{s o l u t e}{s o l v e n t} * 10^{12}$
Density
- $d = \frac{m}{V}$
% Mass >> Molar Concentration
- $\frac{% g S o l u t e}{100 g S o l u t i o n} = \frac{M o l S o l u t e}{V o l S o l u t i o n}$
Examples
- Problem Set - Quantities
  - Nitrogen has the following isotopic abundances:
  - 14.003074 amu 99.6320%
  - 15.000109 amu 0.3680%
  - Determine the molar mass of nitrogen
  - $(14.003074 * 0.996320) + (15.000109 * 0.003680)$
  - $14.0067430888$
  - A compound contains only carbon, hydrogen, and sulphur. A combustion analyzer collects 11.0 g of CO2, 11.26 g of H2O, and 16.0 g of SO2.
  - Determine the empirical formula of the compound.
  - If the molar mass is 98.0 g/mol, determine the molecular formula.
  - $C_{?} H_{?} S_{?} + O_{2} >> C O_{2} + H_{2} O + S O_{2}$
  - $X m o l C = 11.0 g C O_{2} * \frac{1 m o l}{44.01 g} * \frac{1 C}{1 C O_{2}}$
    - $0.2499431947$
    - $0.250$
  - $X m o l H = 11.26 g H_{2} O * \frac{1 m o l}{18.02 g} * \frac{2 H}{1 H_{2} O}$
    - $1.2497225306$
    - $1.250$
  - $X m o l S = 16.0 g S O_{2} * \frac{1 m o l}{64.06 g} * \frac{1 S}{1 S O_{2}}$
    - $0.2497658445$
    - $0.250$
  - $C O_{2} = 44.01 g / m o l$
  - $H_{2} O = 18.02 g / m o l$
  - $S O_{2} = 64.06 g / m o l$
  - $C_{0.250} H_{1.250} S_{0.250}$
  - $C H_{5} S$ = Empirical formula
  - $\frac{98 g / m o l}{(12.011) (5 * 1.008) (32.065)}$
  - $\frac{98 g / m o l}{49.116}$
  - $\frac{98 g / m o l}{49.116}$
  - $1.9952764883$
  - $2$
  - $C_{1 * 2} H_{5 * 2} S_{1 * 2}$
  - $C_{2} H_{10} S_{2}$
  - A 20.0 g sample of MgSO4 . xH2O is dried yielding 13.78 g of anhydrous salt. Determine the formula of the hydrate.
  - $M a s s H_{2} O = 20.0 g - 13.78 g$
  - $M a s s H_{2} O = 6.22 g$
  - $M a s s M g S O_{4} = 13.78 g$
  - $X m o l M g S O_{4} = 13.78 g * \frac{1 m o l}{120.3676 g}$
    - $0.1144826349$
  - $X m o l H_{2} O = 6.22 g * \frac{1 m o l}{18.02 g}$
    - $0.3451720311$
  - $M g S O_{4} = 120.3676 g$
  - $H_{2} O = 18.02 g$
  - $M g S O_{4} + X H_{2} O$
  - $M g S O_{4} + (\frac{0.3451720311}{0.1144826349}) H_{2} O$
  - $M g S O_{4} + 3.0150601565 H_{2} O$
  - $M g S O_{4} + 3 H_{2} O$
  - 10.00 g of copper were mixed with 40.00 g of silver nitrate according to the following reaction:
  - $C u + 2 A g N O_{3} >> 2 A g + C u (N O_{3})_{2}$
  - $90.0 % Y i e l d$
  - How much silver was produced?
  - $X m o l A g = 10.00 g C u * \frac{1 m o l}{63.546 g} * \frac{2 A g}{1 C u}$
  - $X m o l A g = 0.3147326346$
  - $X m o l A g = 0.3147$
  - $X m o l A g = 40.00 g 2 A g N O_{3} * \frac{1 m o l}{339.6 g} * \frac{2 A g}{2 A g N O_{3}}$
  - $X m o l A g = 0.1177856302$
  - $X m o l A g = 0.1178$
  - $∴$ Silver nitrate is the limiting reactant
  - $M a s s A g = 0.1178 (0.900) m o l A g * \frac{107.8682 g}{1 m o l}$
  - $M a s s A g = 11.436186564$