Moles=MassMolar Mass=0.46 g46.0 g mol-1=0.0100 molMoles equals the fraction with numerator Mass and denominator Molar Mass end-fraction equals the fraction with numerator 0.46 g and denominator 46.0 g mol to the negative 1 power end-fraction equals 0.0100 mol Step 3: Calculate the Enthalpy of Combustion (
The following is a representative solution for Task 2 on the Chemsheets worksheet: Brentford School for Girls cap C sub 6 cap H sub 14 water; temperature rose from cap delta cap H (to 3 sig figs). Key Answer Values (Task 2 Summary) Based on Chemsheets marking materials: Question 1: Question 2: Common Sources of Error in Calorimetry
[ \Delta H = \fracqn \quad \text(J mol⁻¹ → usually kJ mol⁻¹) ] ( n ) = moles of limiting reactant or substance being studied.
Using the mass of the reactant rather than the mass of the solution ( calorimetry worksheet 2 answers chemsheets
) represents the heat exchange per mole of reactant under constant pressure, usually expressed in kilojoules per mole ( kJ mol-1kJ mol to the negative 1 power To convert , use this two-step process: Heat in kJ=q1000Heat in kJ equals q over 1000 end-fraction Divide by Moles ( ) of the limiting reactant:
A 50.0 g sample of water at 90.0°C is added to a calorimeter containing 50.0 g of water at 20.0°C. The final temperature of the mixture is 45.0°C. Calculate the heat capacity of the calorimeter. (Specific heat of water = 4.18 J/g°C).
The heat of combustion of sucrose is -20.0 kJ/g or -6846 kJ/mol (Literature value is -5640 kJ/mol, showing the need for calibrated equipment). Moles=MassMolar Mass=0
| Mistake | Correction | |---------|-------------| | Forgetting sign of ( \Delta H ) | Exothermic = negative, endothermic = positive | | Using ( m ) of fuel instead of water | ( m ) = mass of surroundings (water/solution) | | Ignoring heat capacity of calorimeter | If given calorimeter constant ( C ), use ( q = C\Delta T + m_\textwaterc\Delta T ) | | Wrong ( \Delta T ) (e.g., using final only) | ( \Delta T = T_\textfinal - T_\textinitial ) | | Units not converted to kJ | ( \Delta H ) usually in kJ mol⁻¹ → divide J by 1000 |
: Mass of the substance being heated (usually water or the solution). If a solution volume is given, assume : Specific heat capacity (usually for water/solutions). ΔTcap delta cap T : The change in temperature. Calculate Moles (
Add a for exothermic reactions (temperature rise) and a positive sign for endothermic reactions (temperature fall). Common Troubleshooting Tips The final temperature of the mixture is 45
Whether the question involves
ΔH=−qn⋅1000cap delta cap H equals the fraction with numerator negative q and denominator n center dot 1000 end-fraction to convert . The value is negative for exothermic combustion. Common Errors to Note : Use the mass of the water being heated in , not the mass of the fuel.