Relative atomic mass

Relative atomic mass

  • A standard is an element compared with which another element's mass is compared against
    • The standard used is the carbon-12 isotope which is assigned a mass of 12.000 atomic mass units
    • 1 atomic mass unit is 112\frac{1}{12} the mass of one atom of 12C^{12}C
  • Relative atomic mass is unitless relative atomic mass, Ar=average mass of one atom of an element112×the mass of one atom of 12C\text{relative atomic mass, } A_{r} = \frac{\text{average mass of one atom of an element}}{\frac{1}{12}\times \text{the mass of one atom of } ^{12}C}
  • ArA_{r} is calculated based on the weighted average of the relative isotopic masses
Example

for chlorine, the isotopic abundances of its two stable isotopes, 35C^{35} C and 37C^{37} C are around 75%75\% and 25%25\% Ar of Cl=(75100×35)+(25100×37)=35.5A_{r} \text{ of Cl} = \left( \frac{75}{100} \times 35 \right) + \left( \frac{25}{100} \times 37 \right) = 35.5

Relative molecular mass

  • The relative molecular mass of a molecule is the sum of the relative atomic masses of its atoms
  • Relative molecular mass is unitless Relative molecular mass, Mr=average mass of one molecule of a substance112×the mass of one atom of 12C\text{Relative molecular mass, }M_{r} = \frac{\text{average mass of one molecule of a substance}}{\frac{1}{12}\times \text{the mass of one atom of } ^{12} C}
  • For covalent compounds, the relative molecular mass of a substance is calculated from the sum of the relative atomic masses of all the atoms in the formula relative molecular mass, Mr=∑ Ar of all atoms in the molecule\text{relative molecular mass, }M_{r} = \sum \text{ $A_r$ of all atoms in the molecule}
  • For ionic compounds, the term relative formula mass is used instead since they do not exist as molecules relative formula mass, Mr=∑ Ar of all ions in a formula unit\text{relative formula mass, }M_{r} = \sum \text{ $A_r$ of all ions in a formula unit}

Mass spectra

idk dont think need

Percentage composition

  • The percentage composition (percentage by mass) of an element in a compound is the proportion of an element by mass which is present in the compound %composition=no. of atoms of element in compound×Ar of elementMr of compound×100%\text{\%composition} = \frac{\text{no. of atoms of element in compound}\times \text{$A_r$ of element}}{M_{r} \text{ of compound}} \times 100\%
  • Percentage composition can also be used to calculate the proportion of small (discrete) molecules
Example

\ceMgSO4\ce{MgSO4} contains 51.14% of water of crystallisation, deduce formula of hydrated salt

let no. of water of crystallisation be xx x×Mr of \ceH2OMr of \ceMgSO4∗xH2O=51.14100\frac{x\times M_{r} \text{ of }\ce{H2O}}{M_{r}\text{ of }\ce{MgSO4 * xH2O}} = \frac{51.14}{100} thus, formula of hydrated salt is \ceMgSO4∗7H2O\ce{MgSO4 * 7H2O}

Mole

  • The mole is the SI unit of measurement to measure the amount of particles such as atoms, ions, molecules, subatomic particles like electrons, etc
    • The symbol is mol
  • A mole is the amount of substance which contains the same number of particles as the number of atoms of carbon-12
  • 1 mole of substance contains roughly 6.02×10236.02\times 10^{23} particles, known as Avogadro's number
    • Avogadro's constant (NAN_{A}) has a unit of mol number of moles of X, ηx=number of particles in substance XAvogadro’s Constant (NA)\text{number of moles of $X$, }\eta_{x} = \frac{\text{number of particles in substance $X$}}{\text{Avogadro's Constant ($N_A$)}}
  • The molar mass is the mass of one mole of a substance
    • It is numerically equal to the ArA_{r} or MrM_{r} of the substance
    • Its unit is g mol−1g\ mol^{-1} ηX=mass of substance Xmolar mass\eta_{X} = \frac{\text{mass of substance X}}{\text{molar mass}}

Empirical and molecular formula

  • The empirical formula of a compound is the simplest formula which shows the ratio of the atoms of the different elements in the compound
  • When calculating, must be shown as
Element--
Mass of element (g)--
Molar mass of element (g mol−1g\ mol^{-1})--
No. of moles (mol)--
Mole ratio--
Simplest ratio--

Therefore, empirical formula of compound is ___

  • All intermediate steps must be in 4sf
  • All workings must be shown inside the cells of the table
Example

ElementFeO
Mass of each element in 100g of sample (g)21.638.4
Molar mass of element (g mol−1g\ mol^{-1})27.032.1
No. of moles (mol)21.627.0=0.8000(4s.f.)\frac{21.6}{27.0}=0.8000(\text{4s.f.})38.432.1=1.196(4s.f.)\frac{38.4}{32.1}=1.196(\text{4s.f.})
Mole ratio0.8000.800=1\frac{0.800}{0.800} = 11.1960.8=1.495(4s.f.)\frac{1.196}{0.8}=1.495(\text{4s.f.})
Simplest ratio23
The empirical formula is \ceAl2S3\ce{Al2S3}
  • The molecular formula of a compound is the formula which shows the actual number of atoms of each element in one molecule of the compound molecular formula=(empirical formula)nwhere n is an integer\text{molecular formula} = (\text{empirical formula})_{n}\quad \text{where $n$ is an integer}

Molar volume of gases

  • Avogadro's Law
    • Equal volumes of all gases, under the same conditions of temperature and pressure, contain the same number of particles
  • Molar volume
    • The volume of gas occupied by 1 mole of substance
ηgas=vmolar volumewhereηgasin molesvvolume of gas in dm3molar volume of gas at r.t.p.24dm3 mol−1molar volume of gas at s.t.p.22.7dm3 mol−1\begin{array}{c} \begin{aligned} \eta_{\text{gas}} &= \frac{v}{\text{molar volume}} \end{aligned} \\ \hspace{-9em}\textbf{where} \\ \begin{array}{rl} \eta_{\text{gas}} & \text{in moles} \\ v & \text{volume of gas in $dm^3$} \\ \text{molar volume of gas at r.t.p.} & \text{$24dm^3\ mol^{-1}$} \\ \text{molar volume of gas at s.t.p.} & \text{$22.7dm^3\ mol^{-1}$} \end{array} \end{array} 
- One mole of **any** gas at room temperature and pressure (r.t.p.), $24^\circ C$ and $1\text{ atm}=1.01\times 10^5\text{ Pa}$, occupies $24\text{dm}^3$
  • For gases in a balanced equation, volume ratio = mole ratio
    • use for questions to find mole ratio when given volume of gases
Example

20 cm320\text{ cm}^3 hydrocarbon + 100 cm3100\text{ cm}^3 \ceO2\ce{O2} burnt. Remaining gas 70 cm370\text{ cm}^3, reduced to 10 cm310\text{ cm}^3 after being passed through KOH\text{KOH}. Determine hydrocarbon formula (at r.t.p.).

Let formula of hydrocarbon be CxHyC_{x}H_{y}

\ceCxHy(g)\ce{C_xH_y(g)}\ce(x+y4)O2(g)\ce{(x + \frac{y}{4})O2(g)}\cexCO2(g)\ce{xCO2(g)}\cey2H2O\ce{\frac{y}{2}H2O}
initial volume / cm3\text{cm}^320100––
final volume / cm3\text{cm}^301060–
reacting volume / cm3\text{cm}^3209060–
Mole ratio14.53–

x=nCO2nCxHy=31∴x=3x = \frac{n_{\text{CO}_2}}{n_{\text{C}_x\text{H}_y}} = \frac{3}{1} \quad \therefore x = 3

nO2nCxHy=3+y41=4.5∴y=6\frac{n_{\text{O}_2}}{n_{\text{C}_x\text{H}_y}} = \frac{3 + \frac{y}{4}}{1} = 4.5 \quad \therefore y = 6

The hydrocarbon's formula is \ceC3H6\ce{C3H6}

Stoichiometry

  • Stoichiometry refers to the study of the quantitative aspects of chemical formulae and reactions and involves the determination of chemical formulae as well as calculations using balanced chemical equations
    • The stoichiometric coefficient is the coefficient in front of each reactant and product in a balanced equation
  • When doing questions involving mole ratios
    • must write out a statement
    • use 4s.f. for intermediate steps
    • use 1d.p. for molar mass of elements
Example

When heated, calcium carbonate produces calcium oxide and carbon dioxide gas in the following equation: \ceCaCO3(s)−>[heat]CaO(s)+CO2(g)\ce{CaCO3(s) ->[heat] CaO(s) + CO2(g)}

20.5g of \ceCaCO3\ce{CaCO3} was heated

Calculated the mass of calcium oxide produced

\eta_{\ce{CaCO3}} &= \left( \frac{20.5}{40.1+12.0+3(16.0)} \right)\\ &= 0.2048\text{ mol} \\ \text{since } \frac{\eta_{\ce{CaO}}}{\eta_{\ce{CaCO3}}}=\frac{1}{1}\text{, } \eta_{\ce{Cao}} &= 1\times 0.2048\text{ mol} = 0.2048\text{ mol}\\ \therefore \text{ mass of }\ce{CaO} &= 0.2048\text{ mol} \times (40.1 + 16.0)\text{g/mol$^{-1}$} = 11.5\text{g (3s.f.)} \end{align}$$

Limiting reagent

  • A limiting reactant is a reactant that is completely used up in a reaction
  • The other reactants not used up in a reaction are in excess
  • If two or more quantities for reactants are given, limiting reagent must be checked and moles of product must be calculated based on mole ratio to the limiting reagent
  • Must present in this format
    • Write balanced chemical equation
    • Determine moles of each reactant
    • Use the mole ratio from the balanced chemical equation to determine the limiting reagent
    • Choose any reactant and determine the amount required for the other reactant
      • To completely react with __mol of chosen reactant,
      • η\eta of other reactant required = __mol (< or >) of __ mol of other reactant available
      • ∴\therefore __ is in excess
    Example

    Calcium carbide (CaC2CaC_{2}) and carbon monoxide are products of a reaction between calcium oxide and carbon. Calculate the maximum muss of \ceCaC2\ce{CaC2} that can be obtained from 40g of calcium oxide and 40g of carbon.

    • CaO (s)+3C (s)→CaC2 (s)+CO (g)\text{CaO (s)} + 3\text{C (s)} \rightarrow \text{CaC}_2\text{ (s)} + \text{CO (g)}
    • ηCaO=massmolar mass=4040.1+16.0=0.7130 mol\eta_{\text{CaO}} = \frac{\text{mass}}{\text{molar mass}} = \frac{40}{40.1+16.0} = 0.7130\text{ mol}
    • ηC=massmolar mass=4012.0=3.333 mol\eta_{\text{C}} = \frac{\text{mass}}{\text{molar mass}} = \frac{40}{12.0} = 3.333\text{ mol}
    • ηCηCaO=31\frac{\eta_{\text{C}}}{\eta_{\text{CaO}}} = \frac{3}{1}
    • To completely react with 0.7130 mol0.7130\text{ mol} of CaO\text{CaO},
      • \eta_{\text{C}}\text{ required} = 0.7130 \times \frac{3}{1} = 2.139\text{ mol} &lt; 3.333\text{ mol of }\eta_{\text{C}}\text{ available}
    • ∴\therefore Carbon is in excess and CaO\text{CaO} is the limiting reagent.
    • ηCaC2ηCaO=11\frac{\eta_{\text{CaC}_2}}{\eta_{\text{CaO}}} = \frac{1}{1}
    • ∴ηCaC2 formed=11×0.7130=0.7130 mol\therefore \eta_{\text{CaC}_2}\text{ formed} = \frac{1}{1} \times 0.7130 = 0.7130\text{ mol}
    • Molar mass of CaC2=40.1+2(12.0)=64.1 g mol−1\text{Molar mass of CaC}_2 = 40.1 + 2(12.0) = 64.1\text{ g mol}^{-1}
    • ∴Mass of CaC2 formed=0.7130 mol×64.1 g mol−1=45.7 g\therefore \text{Mass of CaC}_2\text{ formed} = 0.7130\text{ mol} \times 64.1\text{ g mol}^{-1} = 45.7\text{ g}

Percentage yield/purity

%yield=experimental mass of producttheoretical mass of product×100%\text{\%yield}= \frac{\text{experimental mass of product}}{\text{theoretical mass of product}} \times 100\% %purity=mass of pure substance presentmass of sample×100%\text{\%purity} = \frac{\text{mass of pure substance present}}{\text{mass of sample}} \times 100\%

Concentration of solutions

  • Solution
    • Solute dissolved in a solvent
  • Concentration
    • Amount of solute dissolved per unit volume of solvent
    • Can be expressed as molar concentration in mol dm3\text{mol}\ \text{dm}^3 [X]=no. of moles of solute X (mol)volume of solution (dm3)[X] = \frac{\text{no. of moles of solute } X \text{ (mol)}}{\text{volume of solution (dm}^3\text{)}}
    • Or concentration in g mol dm3\text{g}\ \text{mol dm}^3 concentration of X (g dm−3)=mass of solute X (g)volume of solution (dm3)\text{concentration of X (g dm}^{-3}\text{)} = \frac{\text{mass of solute } X \text{ (g)}}{\text{volume of solution (dm}^3\text{)}}
    • mol dm−3×molar mass (g mol−1)=g dm−3\text{mol dm}^{-3} \times \text{molar mass (g mol}^{-1}\text{)} = \text{g dm}^{-3}
    • g dm−3÷molar mass (g mol−1)=mol dm−3\text{g dm}^{-3} \div \text{molar mass (g mol}^{-1}\text{)} = \text{mol dm}^{-3}

Dilution

amount of solute=initial concentration×initial volume=final concentration×final volume\text{amount of solute} = \text{initial concentration}\times \text{initial volume} = \text{final concentration} \times \text{final volume}

Back titration

  • A known excess of reagent Y is reacted against an unknown amount of reagent X
  • The amount of unreacted reagent Y is determined using titration with reagent Z of known concentration
  • From this, amount of reacted reagent Y is known, and using stoichiometry, the amount of X is calculated
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