Chapter overview

Overview

Using abundance

  • Nitrogen-14 - 14003074amu - 99.632%
  • Nitrogen-15 - 15.00109amu - 0.0368%
  • \((14.00307\cdot0.9623)+(15.00019 \cdot0.00368)=13.95154+0.0552004\)
  • 14.007amu

Mole

  • Avogadro’s Number: The number of parts of a mole
  • 6.02x1023
  • Used to count things that are small
  • For a element: a mole is the atomic mass in grams
  • For a compound: a mole is the formula mass in grams
  • Molar mass: the mass of a mole of elements, grams Molar map

Molar mas practice

$$1\ mol\ S = 32.065g$$ $$1\ mol\ H_2O = 18.02g$$ $$1\ mol\ NaCl = 58.44g$$ $$1\ mol\ HCl = 58.44g$$ $$1\ mol\ Co = 58.93g$$ $$1\ mol\ KO_{2} = 71.1g$$

Converting from moles to grams

$$ \frac{50.0gC}{1}\times\frac{1\ molC}{12.01gC}=4.16\ mol\ C $$

$$ \frac{1.26molCO_{2}}{1}\times\frac{44.01gCO_{2}}{1molCO_{2}}=55.gCO_{2} $$

The other side of the mole map (formula units)

  • A mole of sulfur is 6.02x1023 atoms$
  • A mole of water is 6.02x1023 molecules$

Examples

$$ 1.204 \times 10^{25} atomFe \times \frac{1\ mole}{6.02 \times 10^{23} atomFe} = 20.0\ mol\ Fe $$

$$ 3.01 \times 10^{20} atom H_{2}0 \times \frac{1mole}{6.02 \times 10^{23} atom H_{2}0} \times \frac{18.015gH_{2}O}{1mole}= 0.00901gH_{2}O $$

  • 0.5012 L of gas to moles of gas $$ 0.5012\ L\ gas \times \frac{1\ mole\ gas}{22.41\ L\ gas}=0.62237\ mole\ gas $$
  • 2.409x23 He atoms to Liters He $$ 2.409e^{ 23 } \times \frac{1molHe}{6.02 \times 10^{23} atomsHe} \times \frac{22.41LHe}{1 molHe}= 8.965\ Liter\ He $$
  • Another $$ 2.91\text{mol NaCl} \times \frac{6.02 \times 10^{23} \text{formula unit}}{1 \text{mole NaCl}} $$
  • 2.91 g CaCO3 to moles CaCO3 $$ 6.27gCaCO_{3} \times \frac{1\ mole\ CaCO_{3}}{100.09g\ CaCO_{3}} = 0.0626\ mole\ CaCO_{3} $$

Stoichiometry

  • Break down the word:
    • Stoichio - “Elements”
    • Metro - “To measure”
  • Mole of moles -> moles of water
  • Road map

Mole to mole

  • 2Al3O3 → 4Al + 3O2

    $$ 6.0\ mol\ Al_{2}O_{3} \times \frac{4\ molAl}{2\ A_{2}O_{3}} = 12molAl $$

  • CO2 + 2LiOH → Li2CO3 + H2O

    $$ 30\ mol\ CO_{2} \times \frac{2\ mol\ LiOH}{1\ mol\ CO_{2}} = 60molLiOH $$

  • N2 + 3H2 → 2NH

    $$ 2.0molH_{2} \times \frac{2\ molNH_{3}}{2\ molH_{2}}=1.33\ molNH_{3} $$

  • 2NaOH + H2SO4 → 2H2O + Na2SO4

    $$ 25molNaOH \times \frac{1H_{2}SO_{4}}{2NoOH}=12.5H_{2}SO_{4} $$

  • 4Fe + 3O2 → 2Fe2O3

$$ 50.0\pu{ mol Fe }\times \frac{2\pu{ mol Fe_{2}O_{3} }}{4\ \pu{ mol Fe }}=25.0\pu{ mol Fe_{2}O_{3} } $$

Mole to mass

  • 6CO2 + 6H2O → 1C6H12O6 + 6O2

  • Given 3.00 mol of water, how much C6H12O6 will be produced

$$ 3.00 molH_{2}O \times \frac{1 mol C_{6}H_{12}O_{6} }{6\ H_{2}O } \times \frac{180g\ C_{6}H_{12}O_{6}}{1\ mol\ C_{6}H_{12}O_{6}}= 90.0gC_{6}H_{12}O_{6} $$

Mass to mole

  • 5C + 2SO2 → 1CS2 + 4CO

  • 8.00g of SO2 mol of CS2

    $$ 8.00g\ SO_{2} \times \frac{1\ mol\ SO_{2}}{64.064g\ SO_{4}} \times \frac{1\ mol\ CS_{2}}{2\ mol\ SO_{2}}=0.06\ mol\ CS_{2} $$

  • If 2.15 mol AgNO3 reacts with NaCl how much NaNO3 is formed in lbs

  • AgNO3 + NaCl → NaNO3 + AgCl

    $$ 2.15molAgNO_{3} \times \frac{1molNaNO_{3}}{1molAgNO_{3}} \times \frac{84.9965gNaNO_{3}}{1molNaNO_{3}} \times \frac{1lb}{454g}=0.403gNaNO_{3} $$

  • Fluorine reacts with sodium cloride to form sodium fluoride and chloride

  • F2 + 2NaCl → 2NaF + Cl2

  • How much sodium fluoride for 45g of fluoride $$ 45gF_{2}\times \frac{1molF_{2}}{37.996gF_{2}} \times \frac{2molNaF}{1molF_{2}}\times \frac{41.988gNaF}{1molNaF}=99.5gNaF $$

Limiting factor

  • CH4 + 2O2 → CO2 + 2H2O

    $$ 32gCH_{4} \times \frac{1molCH_{4}}{16.05gCH_{4}} \times \frac{1molCO_{2}}{1molCH_{4}}=2.0molCO_{2} $$

    $$ 32gO_{2} \times \frac{1molO_{2}}{32.0gO_{2}} \times \frac{1CO_{2}}{2molO_{2}}=0.5molCO_{2} $$

  • Oxygen was the limiting factor with 0.5moles

Percent yield

$$ \text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100% $$

  • Theoretical yield

    • The maximum amount of product that could be produced
  • Actual yield

    • The amount of product that was actually produced when doing the reaction
  • C6H6 + Cl2 → C2H5Cl + HCl

  • 36.8g C6H6 38.8g C6H5Cl

    $$ 38.8gC_{6}H_{6} \times \frac{1molC_{6}H_{5}}{78.06gC_{6}H_{6}} \times \frac{1molC_{2}H_{5}Cl}{1molC_{6}H_{6}} \times \frac{113gC_{2}H_{5}Cl}{1molC_{2}H_{5}Cl}=53.2gC_{6}H_{5}Cl $$

    $$ \frac{38.5g}{53.2g} \times 100 = 75% $$

Percent Composition

  • Percentage of a compound which a element takes up $$ \text{Percent Composition} = \frac{\text{Mass of Element}}{\text{Mass of Compound}} \times 100% $$

Empirical Formula

  • The chemical formula sharing the simplest ratio of elements in a compound
  • C6H12O6
  • CH2O

  • What is the empirical formula for the ingredient in plaster that is 29.44% calcium, 23.56 sulfur, and 47.01% oxygen $$ 29.44gCa\times \frac{1molCa}{40.08gCa}=0.73molCa \div 0.73 = 1 $$ $$ 23.56gS\times \frac{1molS}{32.064gS}=0.73molS \div 0.73 = 1 $$ $$ 47.01gO \times \frac{1molO}{16gO}=2.94molO \div 0.73 = 4 $$
  • CaSO4 (calcium sulfate)
  • Another
  • 22.0% sulpher, 78.0% cadiumium $$ 22.0gS\times \frac{1molS}{32.064gS}=0.686g \div 0.686 = 1 $$ $$ 78.0gCd\times \frac{1molCd}{112.41gCd}=0.693g \div 0.686 = 1.01 $$
  • CdS

Molecular Formula

  • A compound has a molar mass of 26.04g/mog and a empirical formula of CH. What is the molecular formula?
  • We went from C6H2O6 to CH2O. Now we are working backwards $$ \frac{26.04}{13.02}=2 $$ $$ 2(CH) = C_{2}H_{2} $$
  • Empirical formula P2O5 molar mass 283.89 $$ \frac{283.89}{141.94} = 2 $$ $$ 2(P_{2}O_{5})=P_{4}O_{10} $$
  • The formula is $$ \frac{\text{Whole Compound}}{\text{Simplified Compound}} $$
  • A compound has 14.87% phosphorus, 85.13% chlorine. Molar mass 208.239g what formula $$ 13.87gP \times \frac{1molP}{30.97gP}=0.48molP \div 0.48 = 1 $$ $$ 85.13gCl \times \frac{1molCl}{35.453gCl}=2.40molCl \div 0.48 = 5 $$
  • PCL5 (were not done!) $$ \frac{208.239}{208.235}=1 $$ $$ 1(PCl_{5})=PCl_{5} $$