Question #2040e
1 Answer
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
Let's assume that you start with
After you partially dry the sample, the mas of water will decrease by an unknown amount
The new sample is now
#(12 - x)/(100 - x) xx 100 = 7%"water"#
Solve this equation for
#(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
#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
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.