Question #2040e

1 Answer
Oct 18, 2015

Answer:

#"47.3%"#

Explanation:

The trick here is to use the fact that the only the amount of water actually changes after the original sample is partially dried.

You can predict that the initial percentage of silica in the sample was smaller than #50%#, since the sample initially contained more water.

Let's assume that you start with #"100 g"# of the original sample. You know that this sample contains #12%# water, which is equivalent to #"12 g"# of water.

After you partially dry the sample, the mas of water will decrease by an unknown amount #x#. In addition to the mass of water, the mass of the sample will also decrease by an amount #x#.

The new sample is now #7%# water, which means that you can write

#(12 - x)/(100 - x) xx 100 = 7%"water"#

Solve this equation for #x# to get

#(12 - x) * 100 = 7 * (100 - x)#

#1200 - 100x = 700 - 7x#

#93x = 500 implies x = 500/93 = "5.376 g"#

The mass of the sample after the water was dried out is

#m_"final" = "100 g" - "5.376 g" = "94.624 g"#

You know that this sample is #50%# silica, which means that you can determine exactly how much silica it contains

#94.624color(red)(cancel(color(black)("g sample"))) * "50 g silica"/(100color(red)(cancel(color(black)("g sample")))) = "47.312 g silica"#

Since drying the sample only reduced the amount of water it contained, it follows that this much silica was a prt of the original #"100-g"# sample as well.

Therefore, the initial percent composition of silica in the sample was

#(47.312color(red)(cancel(color(black)("g"))))/(100color(red)(cancel(color(black)("g")))) xx 100 = color(green)("47.3%")#

I'll leave the answer rounded to three sig figs, despite the fact that the percentages you gave would not justify this.