Do you have a sample calculation for chemistry Experiment 9.2?


Here is a sample calculation:

Data:

Mass of 50.0 mL vinegar: 51.0 g
Number of mL of vinegar to reach endpoint: 23.0 mL

Calculations:

Density of vinegar:

51.0 g/ 50.0 mL = 1.02 g/mL

Number of grams of vinegar added:

(1.02 g/mL) x (23.0 mL) = 23.5 g

Mass of C2H4O2 added:

(23.5 g) x (0.0500) = 1.18 g

Molecular Mass of C2H4O2 = 60.0 amu

Number of moles of C2H4O2 added:

(1.18 g/1) x (1 mole/60.0 grams) = 0.0197 moles

Number of moles of ammonia that exist in solution:

(0.0197 moles C2H4O2 /1)x(1 mole ammonia/1 mole C2H4O2) = 
0.0197 moles ammonia

Concentration of ammonia:

0.0197 moles ammonia / 0.0100 L = 1.97 M

Tags: Chemistry
Last update:
2019-01-29 21:19
Author:
Jacey
Revision:
1.2
Average rating:0 (0 Votes)

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