# 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 *also decrease* by an 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.