1) Welcome to PF
2) Generally, you would post such questions
here. Also, you can sometimes ask PF moderators/admins to 'move' your thread to a more appropriate forum.
3) Your question can easily be solved via a system of linear equations, (although I'm not sure if that's what the
AP Chemistry Exam prefers...). Letting m
X denote numerically the molar mass of the element 'X', the
relative quantities of moles NaCl (x), Na
2SO
4 (y), NaNO
3 (z) can be calculated as the solution to
\begin{gathered}<br />
m_{{\text{Na}}} \left( {x + 2y + z} \right) = 32.08 \hfill \\<br />
m_{\text{O}} \left( {4y + 3z} \right) = 36.01 \hfill \\<br />
m_{{\text{Cl}}} x = 19.51 \hfill \\ <br />
\end{gathered}
Note that we can substitute x=19.51/m
Cl into the 2
nd and 3
rd equations,
and also (for convenience) rewrite them as:
\begin{gathered}<br />
2y + z = 32.08m_{{\text{Na}}}^{ - 1} - 19.51m_{{\text{Cl}}}^{ - 1} \hfill \\ 4y + 3z = 36.01m_{{\text{O}}}^{ - 1} \hfill \\ <br />
\end{gathered}
Elimination (of these two equations) will allow us to immediately solve for both y and z, as
\begin{gathered}<br />
y = \frac{{3\left( {32.08m_{{\text{Na}}}^{ - 1} - 19.51m_{{\text{Cl}}}^{ - 1} } \right) - 36.01m_{\text{O}}^{ - 1} }}<br />
{{\text{2}}} \hfill \\<br />
z = 36.01m_{\text{O}}^{ - 1} - 2\left( {32.08m_{{\text{Na}}}^{ - 1} - 19.51m_{{\text{Cl}}}^{ - 1} } \right) \hfill \\ <br />
\end{gathered}
"Plugging in" the molar masses values for Na, O, and Cl, we find that
x=0.5504
y=0.1414
z=0.5618
To find the mass percents of NaCl, Na
2SO
4, and NaNO
3, simply multiply the molar masses
of each compound by the relative molar quantities of each compound. Thus,
*Mass percent NaCl = 0.5504(58.44) =
32.17%
*Mass percent Na
2SO
4 = 0.1414(142.0) =
20.08%
*Mass percent NaNO
3 = 0.5618(84.99) =
47.75%
-----------------------------------------------------------------------------
Edit: You can check your answers for (x,y,z) on a graphing calculator via solving the
(augmented) matrix:
\left[ {\begin{array}{*{20}c}<br />
\hfill 1 & \hfill 2 & \hfill 1 & \hfill {\text{1.395}}} \\<br />
\hfill 0 & \hfill 4 & \hfill 3 & \hfill {2.251} \\<br />
\hfill 1 & \hfill 0 & \hfill 0 & \hfill {0.5504} \\<br />
\end{array} } \right]
IIRC, the AP Chemistry Exam allows graphing calculators on the free response section, so you should have no trouble with these types of questions.
However, remember that graphing calculators do not usually consider significant figures when solving matrices.