Chemistry Ia
Data Collection: Mass of FA1 | Mass of empty weighing bottle and FA1 (±0.001) /g | 5.821 | Mass of empty weighing bottle (±0.001) /g | 4.321 | Mass of FA1 (±0.002) /g | 1.500 |
Volume of FA3 prepared = 250.00 ± 0.12 cm3
Volume of FA3 used for titration = 20.00 ± 0.06 cm3 Titration of FA3 against FA2 | | Run 1 | Run 2 | Run 3 | Final volume of FA2 (± 0.05) /cm3 | 25.30 | 25.40 | 25.45 | Initial volume of FA2 (± 0.05) /cm3 | 0.00 | 0.00 | 0.00 | Volume of FA2 used (± 0.10) /cm3 | 25.30 | 25.40 | 25.45 | Titration used | | √ | √ |
Qualitative observations:
- FA2 and FA3 are colourless solutions.
- Titration is done until the solution turns permanently pink. Calculations | Propagation of uncertainties | | | Average volume of FA2 used | ± ∆ Average volume of FA2 used | = (25.40 + 25.45)/2 | = ± ½ (0.10+0.10) | = 25.43 cm3 | = ± 0.10 cm3 | | | Average volume of FA2 used = (25.43 ± 0.10) cm3 |
Data processing: Calculations | Propagation of uncertainties | | | Amount of FA1
= Mass of FA1Molar mass of FA1
= 1.5002*12.01+ 6*1.01+ 6*16.00
= 0.011897 molAmount of FA3 = Amount of FA1Concentration of FA3= Amount of FA3 / Vol of FA3 solution= 0.011987 / (250/1000)= 0.047589 mol dm-3Amount of FA3 in 20.0cm3 of FA3= 0.047589 x 20.00/1000= 0.0009518 molMole ratio of FA3:FA2 used is 1:2Amount of FA2 used= 2 x 0.0009518= 0.0019036 molConcentration of FA2=0.0019036 / (25.43/1000)= 0.074856 mol dm-3 | ± % ∆ Amount of FA1
= ± % ∆ Mass of FA1
=± 0.002/1.500 x 100%
= ± 0.13%± % ∆ Amount of FA3
= ± % ∆ Amount of FA1
= ± 0.13%± % ∆ Concentration of FA3= ± % ∆ Amount of FA3 + ± % ∆ Volume of FA3 solution= ± [0.13% + (0.12/250.00 x 100%)= ± 0.18%± % ∆ Amount of FA3 in 20.0cm3 of FA3= ± [0.18% + (0.06/20.00 x 100)%]= ± 0.48%± % ∆ Amount of FA2 used= ± % ∆ Amount of FA3 in 20.0cm3 of FA3= ± 0.48%± % ∆ Concentration of FA2= ± (% ∆ Amount of FA2 used + % ∆ Average volume of FA2 used)= ± [ 0.48% + (0.10/25.43 x 100%)]= ± 0.87% % ∆
Volume of FA3 prepared = 250.00 ± 0.12 cm3
Volume of FA3 used for titration = 20.00 ± 0.06 cm3 Titration of FA3 against FA2 | | Run 1 | Run 2 | Run 3 | Final volume of FA2 (± 0.05) /cm3 | 25.30 | 25.40 | 25.45 | Initial volume of FA2 (± 0.05) /cm3 | 0.00 | 0.00 | 0.00 | Volume of FA2 used (± 0.10) /cm3 | 25.30 | 25.40 | 25.45 | Titration used | | √ | √ |
Qualitative observations:
- FA2 and FA3 are colourless solutions.
- Titration is done until the solution turns permanently pink. Calculations | Propagation of uncertainties | | | Average volume of FA2 used | ± ∆ Average volume of FA2 used | = (25.40 + 25.45)/2 | = ± ½ (0.10+0.10) | = 25.43 cm3 | = ± 0.10 cm3 | | | Average volume of FA2 used = (25.43 ± 0.10) cm3 |
Data processing: Calculations | Propagation of uncertainties | | | Amount of FA1
= Mass of FA1Molar mass of FA1
= 1.5002*12.01+ 6*1.01+ 6*16.00
= 0.011897 molAmount of FA3 = Amount of FA1Concentration of FA3= Amount of FA3 / Vol of FA3 solution= 0.011987 / (250/1000)= 0.047589 mol dm-3Amount of FA3 in 20.0cm3 of FA3= 0.047589 x 20.00/1000= 0.0009518 molMole ratio of FA3:FA2 used is 1:2Amount of FA2 used= 2 x 0.0009518= 0.0019036 molConcentration of FA2=0.0019036 / (25.43/1000)= 0.074856 mol dm-3 | ± % ∆ Amount of FA1
= ± % ∆ Mass of FA1
=± 0.002/1.500 x 100%
= ± 0.13%± % ∆ Amount of FA3
= ± % ∆ Amount of FA1
= ± 0.13%± % ∆ Concentration of FA3= ± % ∆ Amount of FA3 + ± % ∆ Volume of FA3 solution= ± [0.13% + (0.12/250.00 x 100%)= ± 0.18%± % ∆ Amount of FA3 in 20.0cm3 of FA3= ± [0.18% + (0.06/20.00 x 100)%]= ± 0.48%± % ∆ Amount of FA2 used= ± % ∆ Amount of FA3 in 20.0cm3 of FA3= ± 0.48%± % ∆ Concentration of FA2= ± (% ∆ Amount of FA2 used + % ∆ Average volume of FA2 used)= ± [ 0.48% + (0.10/25.43 x 100%)]= ± 0.87% % ∆