An old coin found in an ancient temple is composed of zinc coated with copper. In an experiment to find the percent zinc in the coin, a student determined the weight of the coin to be 3.0 g. Then the student made several scratches in the copper coating (to expose the underlying zinc) and put the scratched coin in hydrochloric acid, where the following reaction occurred between the zinc and HCl (copper remained undissolved):

Zn(s) + 2 HCl (aq) → H2(g) + ZnCl2 (aq)

The student collected the hydrogen produced over water at 27 °C. The collected gas occupied a volume of 1 L at a total pressure of 1.02 bar. The percent zinc in the coin, the student found would be. (Assume that all the Zn in the coin dissolves.)

( a) 67.7 %

(b) 98.2 %

(c) 96.7 %

(d) 88.3 %

(e) 25.0 %

My attempt:

P = 1.02 x 10^5 Pa

V = 1 x 10^-3 m^3

T = 27 + 273 = 300K

Use PV=nRT,

(1.02 x 10^5)(1 x 10^-3)=n x .31 x 300

n = 0.04091 moles of H2

moles of Zn = 0.04091

Mass of Zn = 0.04091 x 65.39 = 2.67 g

Percentage of Zn = (2.67/3.0) x 100 = 89.2%

The mark scheme answer is D) 88.3% which is the closest answer to the value I got, but why don't I get that answer? Is it an error in my calculation, the mcq answers or something else?

  • $\begingroup$ I do not know how advanced the course is, are you supposed to correct for vapor pressure of water? $\endgroup$
    – ACR
    Commented Feb 12, 2023 at 23:35
  • $\begingroup$ This is from a College Preliminary test. Apparently requires Advanced Level Chemistry knowledge only. $\endgroup$
    – Jane902
    Commented Feb 13, 2023 at 0:59
  • $\begingroup$ With considered water vapor, the true hydrogen volume, zinc mass and percentage would be lower, hopefully close to the given answer. $\endgroup$
    – Poutnik
    Commented Feb 15, 2023 at 20:33
  • $\begingroup$ I'd really like to hear from an archaeologist how likely such a scenario would actually be. $\endgroup$ Commented Feb 19, 2023 at 23:47

1 Answer 1


If the original data gives an original weight of $3.0$ g, it means that the real weight is situated between $2.95$ g and $3.05$ g. It also means that the precision on this weight is $±0.05$ g, which correspond to ± $0.05/3.0 = 1.7$%. The final result is also known with a precision of ±$1.7$%. And ±$1.7$% of $88.3$ is ±$1.47$ Take care of the following calculations. Don't confuse percent and percent of percent. The final value is something between $88,3 - 1.47 = 86.8$% and $88.3 + 1.47 = 89.7$%. As a consequence, your result ($89.2$%) is within these limits. It is valid.

  • 1
    $\begingroup$ Rules of thumb such as "3.0 g, it means that the real weight is situated between 2.95 g and 3.05 g. It also means that the precision on this weight is ±0.05 g" are not valid in real world analytical chemistry. $\endgroup$
    – ACR
    Commented Feb 12, 2023 at 16:46

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