Science- Chemistry

posted by .

A test tube containing a mixture of compounds X, Y, and a small amount of Z weighs 25.4886 g. The tube is emptied into a beaker and the empty tube is reweighed, giving a value of 10.2586 g. The mixture is separated into two components by dissolving in water, filtering off the insoluble material (Component X) and then crystallizing one component (Y) from the filtrate. The following data was obtained:

Mass of filter paper/Lg.watch glass: 12.9912 g
Mass of filter paper/Lg.watch glass + X: 23.5899 g
Mass of filter paper/Sm.watch glass: 9.4332 g
Mass of filter paper/Sm.watch glass + Y: 12.1622 g

Calculate the following: 1. Mass of sample
2. Mass of X recovered
3. Mass of Y recovered
4. % X in the sample
5. % Y in the sample
6. Assuming the actual composition is 69.00% X and 31.00% Y, calculate the % error for each component. Indicate if the %Error represents a HIGH or LOW error relative to the accepted values.

• Science- Chemistry -

Is this right?

1. Mass of Sample:
= (mass of total)- (mass of empty tube)
= 25.4886 g - 10.2586 g
= 15.2300 g
2. Mass of X recovered:
= (mass of filter paper/large watch glass + X) - (mass of filter paper/lg. watch glass)
= 23.5899 g - 12.9912 g
= 10.5987 g
3. Mass of Y recovered:
= (mass of filter paper/small watch glass + Y) - (mass of filter paper/sm. watch glass)
= 12.1622 g - 9.4332 g
= 2.7290 g
4. % of X in sample:
= [(mass of X) / (mass of total)] * 100%
= (10.5987 g / 25.4886 g) * 100%
= 41.6%
5. % of Y in sample:
= [(mass of Y) / (mass of total)] * 100%
= (2.7290 g / 25.4886 g) * 100%
= 10.7%
6. Percent Error:
= [(|accepted value - determined value|) / (accepted value)] * 100%
For X:
= [(|.6900 - .416|) / (.6900)] * 100%
= 39.7%
For Y:
= [(|.3100 - .107|) / (.3100)] * 100%
= 65.4%