Separation of a Mixture of Solids

Topics: Separation process, Mixture, Separation Pages: 4 (618 words) Published: July 1, 2013
Separation of a Mixture of Solids

Lab: Separation of a Mixture of Solids Experiment
 

PURPOSE
The purpose of this experiment was to learn how to separate a mixture of solids.

DATA

| | |Data Table 1: Experiment Data | | |Grams |Percent of mixture | |Iron filings |2.2 |30.56 | |Sand |1.3 |18.06 | |Table salt |1.5 |20.83 | |Benzoic acid |1.1 |15.28 | |Total |6.1 |84.73 |

Net mass of mixture:
Mass of weighing dish: 0.6g
Total of mixture and weighing dish: 7.8g

7.8g – 0.6g = 7.2g

Net mass of iron fillings:
Square piece of paper: 0.7g
Paper and Iron fillings: 2.9g

2.9g – 0.7g = 2.2g

Net mass of sand:
Weighing dish and sand: 1.9g
Mass of weighing dish: 0.6g

1.9g – 0.6g = 1.3g

Net mass of benzoic acid:
Filter paper: 1.0g
Filter Paper and Benzoic acid 2.1g

2.1g – 1.0g = 1.1g

Net mass of salt:
Salt + cup: 6.2g
Cup: 4.7g

6.2g – 4.7g = 1.5g

Total net mass of samples collected:
2.2g + 1.3g + 1.5g +1.1g = 6.1g

Percentage of total mixture:
(Calculated using initial net mass of sample, 7.2g)
2.2g / 7.2g = 30.56%
1.3g / 7.2g = 18.06%
1.5g / 7.2g = 20.83%
1.1g / 7.2g = 15.28%

Percent error:
((6.1g -7.2g) / (7.2g)) * 100 = 15.28%

OBSERVATIONS
During the heating of the sand, I forgot to cover the top of the beaker with a saucer. This resulted in a loss of sample, which impacted my results.

CONCLUSION/DISCUSSION

When I calculated the percent error of the initial total net mass of the mixture...
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