IB — chemistryMedium3 min read

Stoichiometric Relationships

Master the mole concept, empirical and molecular formulae, and solution stoichiometry for IB Chemistry.

Tutor AI Team
Tutor AI Team

# Stoichiometric Relationships

Stoichiometry is the quantitative study of chemical reactions — calculating how much reacts and how much is produced. It forms the backbone of IB Chemistry and connects to virtually every other topic.

1. The Mole Concept

The mole is the SI unit for amount of substance. One mole contains exactly 6.022×10236.022 \times 10^{23} particles (Avogadro's constant, NAN_A).

n=mMn = \frac{m}{M}

where nn = moles, mm = mass (g), MM = molar mass (g/mol)

Molar Volume of Gas

At STP (0°C, 100 kPa): Vm=22.7V_m = 22.7 dm³/mol

n=VVmn = \frac{V}{V_m}

Ideal Gas Law

PV=nRTPV = nRT

where R=8.314R = 8.314 J K⁻¹ mol⁻¹, TT in K, PP in Pa, VV in m³

2. Empirical and Molecular Formulae

Empirical formula: simplest whole-number ratio of atoms

Molecular formula: actual number of atoms in one molecule

Method for Empirical Formula

  1. Divide mass (or %) of each element by its ArA_r
  2. Divide all results by the smallest
  3. Round to nearest whole number (or multiply to get whole numbers)

Finding Molecular Formula

n=Mr(molecular)Mr(empirical)n = \frac{M_r(\text{molecular})}{M_r(\text{empirical})}

Multiply empirical formula subscripts by nn.

Example: Empirical = CH₂O (Mr=30M_r = 30), molecular mass = 180 n=180/30=6→Molecular formula=C6H12O6n = 180/30 = 6 \rightarrow \text{Molecular formula} = \text{C}_6\text{H}_{12}\text{O}_6

3. Balanced Equations and Stoichiometry

Reacting Mass Calculations

  1. Write balanced equation
  2. Calculate moles of known substance
  3. Use molar ratio to find moles of unknown
  4. Convert to mass/volume as needed

Limiting Reagent

The reactant that runs out first determines the maximum product.

Theoretical and Percentage Yield

% yield=actual yieldtheoretical yield×100\text{\% yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100

4. Solution Stoichiometry

c=nVc = \frac{n}{V}

where cc = concentration (mol/dm³), VV = volume (dm³)

Dilution

c1V1=c2V2c_1V_1 = c_2V_2

Titration Calculations

  1. Calculate moles of known solution: n=cVn = cV
  2. Use molar ratio
  3. Find unknown concentration

Worked Example: Example 1

Problem

Question: What mass of CO₂ is produced when 10.0 g of CaCO₃ reacts with excess HCl?

CaCO3+2HCl→CaCl2+H2O+CO2\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2

n(CaCO3)=10.0/100.1=0.0999n(\text{CaCO}_3) = 10.0/100.1 = 0.0999 mol

Ratio 1:1 → n(CO2)=0.0999n(\text{CO}_2) = 0.0999 mol

m(CO2)=0.0999×44.0=4.40m(\text{CO}_2) = 0.0999 \times 44.0 = 4.40 g

Solution

Worked Example: Example 2

Problem

Question: 25.0 cm³ of 0.100 mol/dm³ NaOH requires 20.0 cm³ of H₂SO₄. Find the concentration of H₂SO₄.

2NaOH+H2SO4→Na2SO4+2H2O2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}

n(NaOH)=0.100×0.0250=0.00250n(\text{NaOH}) = 0.100 \times 0.0250 = 0.00250 mol

n(H2SO4)=0.00250/2=0.00125n(\text{H}_2\text{SO}_4) = 0.00250/2 = 0.00125 mol

c=0.00125/0.0200=0.0625c = 0.00125/0.0200 = 0.0625 mol/dm³

Solution

6. Practice Questions

    1. Calculate the number of molecules in 3.6 g of water.
    1. Find the empirical formula of a compound with 40.0% C, 6.7% H, 53.3% O.
    1. What volume of O₂ at STP is needed to completely combust 5.0 g of ethanol?
    1. 2.00 g of Mg reacts with 100 cm³ of 1.00 mol/dm³ HCl. Which is limiting?
    1. A student obtains 8.5 g of product when the theoretical yield is 12.0 g. Calculate percentage yield.

Want to check your answers and get step-by-step solutions?

Get it on Google PlayDownload on the App Store

7. IB Exam Tips

  • Always use the IB Data Booklet for ArA_r values
  • Show all working — marks are given for method even if the final answer is wrong
  • Convert volumes to dm³ before using c=n/Vc = n/V
  • Gas calculations: use PV=nRTPV = nRT at non-standard conditions
  • Significant figures: match the data given (usually 3 s.f.)

Summary

  • n=m/Mn = m/M (mass) = c×Vc \times V (solution) = V/VmV/V_m (gas at STP) = PV/RTPV/RT (ideal gas)
  • Empirical formula from composition; molecular formula from molar mass
  • Limiting reagent determines maximum product
  • % yield = actual/theoretical × 100
Tags:
#ib#chemistry#stoichiometry#moles

Related Topics

Ready to Ace Your IB chemistry?

Get instant step-by-step solutions to any problem. Snap a photo and learn with Tutor AI — your personal exam prep companion.

Get it on Google PlayDownload on the App Store