Chemistry by Raymond Chang - Problem 2.130

Posted on 28 March, 2025

A sample containing NaCl\text{NaCl}, Na2SO4\text{Na}_2\text{SO}_4 and NaNO3\text{NaNO}_3 gives the following elemental analysis: Na 32.08%, O 36.01% and Cl 19.51%. Calculate the mass percent of each compound in the sample.

The relevant molar masses are as follows.

  • Na\text{Na} - 22.99g/mol22.99\text{g}/\text{mol}
  • Cl\text{Cl} - 35.45g/mol35.45\text{g}/\text{mol}
  • S\text{S} - 32.07g/mol32.07\text{g}/\text{mol}
  • O\text{O} - 16.00g/mol16.00\text{g}/\text{mol}
  • N\text{N} - 14.01g/mol14.01\text{g}/\text{mol}

We assume a 100g100\text{g} sample and can then calculate the mass of NaCl\text{NaCl} as the ratio of Cl\text{Cl} to Na\text{Na} is 1:1, only one compound contains Cl\text{Cl} and the mass of Cl\text{Cl} is known.

19.51g×1molCl35.45g×1molNaCl1molCl×22.99g+35.45g1molNaCl 19.51 \text{g} \times \frac{1 \text{mol}_{\text{Cl}}}{35.45 \text{g}} \times \frac{1 \text{mol}_{\text{NaCl}}}{1 \text{mol}_{\text{Cl}}} \times \frac{22.99 \text{g} + 35.45 \text{g}}{1 \text{mol}_{\text{NaCl}}}

=0.5504molCl×1molNaCl1molCl×22.99g+35.45g1molNaCl = 0.5504 \text{mol}_{\text{Cl}} \times \frac{1 \text{mol}_{\text{NaCl}}}{1 \text{mol}_{\text{Cl}}} \times \frac{22.99 \text{g} + 35.45 \text{g}}{1 \text{mol}_{\text{NaCl}}}

=32.17g = 32.17 \text{g}

The mass of the remaining two compounds depend on each other. They each share the remaining Na\text{Na} as well as the 36.01g36.01 \text{g} of O\text{O}.

(Na2SO4)g=(22.99×2+32.07+16.00×4)g1molNa2SO4×(36.01g−(NaNO3)g×16.00g×322.99g+14.01g+16.00g×3)×1molO16.00g×1molNa2SO44molO (\text{Na}_2\text{SO}_4)\text{g} = \frac {(22.99 \times 2 + 32.07 + 16.00 \times 4) \text{g}} {1 \text{mol}_{\text{Na}_2\text{SO}_4}} \times ( 36.01 \text{g} - (\text{NaNO}_3)\text{g} \times \frac {16.00 \text{g} \times 3} { 22.99 \text{g} + 14.01 \text{g} + 16.00 \text{g} \times 3 } ) \times \frac{1 \text{mol}_{\text{O}}}{16.00 \text{g}} \times \frac{1 \text{mol}_{\text{Na}_2\text{SO}_4}}{4 \text{mol}_{\text{O}}}

=142.1g1molNa2SO4×(36.01g−(NaNO3)g×16.00g×385g)×1molO16.00g×1molNa2SO44molO = \frac{142.1 \text{g}}{1 \text{mol}_{\text{Na}_2\text{SO}_4}} \times ( 36.01 \text{g} - (\text{NaNO}_3)\text{g} \times \frac {16.00 \text{g} \times 3} {85 \text{g}} ) \times \frac{1 \text{mol}_{\text{O}}}{16.00 \text{g}} \times \frac{1 \text{mol}_{\text{Na}_2\text{SO}_4}}{4 \text{mol}_{\text{O}}}

=142.1g1molNa2SO4×(36.01g−(NaNO3)g×0.5647)×1molO16.00g×1molNa2SO44molO = \frac{142.1 \text{g}}{1 \text{mol}_{\text{Na}_2\text{SO}_4}} \times (36.01 \text{g} - (\text{NaNO}_3)\text{g} \times 0.5647) \times \frac{1 \text{mol}_{\text{O}}}{16.00 \text{g}} \times \frac{1 \text{mol}_{\text{Na}_2\text{SO}_4}}{4 \text{mol}_{\text{O}}}

=142.1g×(molNa2SO4)−1×(36.01×g−(NaNO3)×g×0.5647)×molO×16.00−1×g−1×molNa2SO4×4−1×(molO)−1 = 142.1 \text{g} \times (\text{mol}_{\text{Na}_2\text{SO}_4})^{-1} \times ( 36.01 \times \text{g} - (\text{NaNO}_3) \times \text{g} \times 0.5647 ) \times \text{mol}_{\text{O}} \times 16.00^{-1} \times \text{g}^{-1} \times \text{mol}_{\text{Na}_2\text{SO}_4} \times 4^{-1} \times (\text{mol}_{\text{O}})^{-1}

=142.1×16.00−1×4−1×(36.01×g−(NaNO3)×g×0.5647)×g = 142.1 \times 16.00^{-1} \times 4^{-1} \times ( 36.01 \times \text{g} - (\text{NaNO}_3) \times \text{g} \times 0.5647 ) \times \text{g}

=142.1×64.00−1×(36.01×g−(NaNO3)×g×0.5647)×g = 142.1 \times 64.00^{-1} \times ( 36.01 \times \text{g} - (\text{NaNO}_3) \times \text{g} \times 0.5647 ) \times \text{g}

=2.220×(36.01g−(NaNO3)×g×0.5647)×g = 2.220 \times (36.01 \text{g} - (\text{NaNO}_3) \times \text{g} \times 0.5647) \times \text{g}

=2.220×36.01×g−2.220×(NaNO3)×0.5647×g = 2.220 \times 36.01 \times \text{g} - 2.220 \times (\text{NaNO}_3) \times 0.5647 \times \text{g}

=79.94g−1.254×(NaNO3)g = 79.94\text{g} - 1.254 \times (\text{NaNO}_3) \text{g}

From this we get an equation for (Na2SO4)g(\text{Na}_2\text{SO}_4)\text{g} which depends on (NaNO3)g(\text{NaNO}_3) \text{g}.

(Na2SO4)g=79.94g−1.254×(NaNO3)g (\text{Na}_2\text{SO}_4)\text{g} = 79.94\text{g} - 1.254 \times (\text{NaNO}_3) \text{g}

We can calculate an alternative equation for (Na2SO4)g(\text{Na}_2\text{SO}_4)\text{g}, referring instead to the remaining Na\text{Na}.

To simplyfy things a little we can calculate the Na\text{Na} remaining after NaCl\text{NaCl} has been accounted for.

32.08g−32.17g×22.99g22.99g+35.45g 32.08 \text{g} - 32.17 \text{g} \times \frac{22.99 \text{g}}{22.99 \text{g} + 35.45\text{g}}

=32.08g−32.17g×0.3934 = 32.08 \text{g} - 32.17 \text{g} \times 0.3934

=32.08g−12.66g = 32.08 \text{g} - 12.66 \text{g}

=19.42g = 19.42 \text{g}

Which we can now use.

(Na2SO4)g=142.1g1molNa2SO4×(19.42g−(NaNO3)g×22.99g22.99g+14.01g+16.00g×3)×1molNa22.99g×1molNa2SO42molNa (\text{Na}_2\text{SO}_4)\text{g} = \frac{142.1 \text{g}}{1 \text{mol}_{\text{Na}_2\text{SO}_4}} \times ( 19.42 \text{g} - (\text{NaNO}_3)\text{g} \times \frac {22.99 \text{g}} { 22.99 \text{g} + 14.01 \text{g} + 16.00 \text{g} \times 3 } ) \times \frac{1 \text{mol}_{\text{Na}}}{22.99 \text{g}} \times \frac {1 \text{mol}_{\text{Na}_2\text{SO}_4}} {2 \text{mol}_{\text{Na}}}

(Na2SO4)g=142.1g1molNa2SO4×(19.42g−(NaNO3)g×22.99g85g)×1molNa22.99g×1molNa2SO42molNa (\text{Na}_2\text{SO}_4)\text{g} = \frac{142.1 \text{g}}{1 \text{mol}_{\text{Na}_2\text{SO}_4}} \times ( 19.42 \text{g} - (\text{NaNO}_3)\text{g} \times \frac{22.99 \text{g}}{85 \text{g}} ) \times \frac{1 \text{mol}_{\text{Na}}}{22.99 \text{g}} \times \frac {1 \text{mol}_{\text{Na}_2\text{SO}_4}} {2 \text{mol}_{\text{Na}}}

=142.1g1molNa2SO4×(19.42g−(NaNO3)g×0.2705)×1molNa22.99g×1molNa2SO42molNa = \frac{142.1 \text{g}}{1 \text{mol}_{\text{Na}_2\text{SO}_4}} \times ( 19.42 \text{g} - (\text{NaNO}_3)\text{g} \times 0.2705 ) \times \frac{1 \text{mol}_{\text{Na}}}{22.99 \text{g}} \times \frac {1 \text{mol}_{\text{Na}_2\text{SO}_4}} {2 \text{mol}_{\text{Na}}}

=142.1×g×(molNa2SO4)−1×(19.42g−(NaNO3)g×0.2705)×molNa×22.99−1×g−1×molNa2SO4×2−1×(molNa)−1 = 142.1 \times \text{g} \times (\text{mol}_{\text{Na}_2\text{SO}_4})^{-1} \times (19.42 \text{g} - (\text{NaNO}_3)\text{g} \times 0.2705) \times \text{mol}_{\text{Na}} \times 22.99^{-1} \times \text{g}^{-1} \times \text{mol}_{\text{Na}_2\text{SO}_4} \times 2^{-1} \times (\text{mol}_{\text{Na}})^{-1}

=142.1×22.99−1×2−1×(19.42g−(NaNO3)g×0.2705)×g×g−1×molNa2SO4×(molNa2SO4)−1×molNa×(molNa)−1 = 142.1 \times 22.99^{-1} \times 2^{-1} \times (19.42 \text{g} - (\text{NaNO}_3)\text{g} \times 0.2705) \times \text{g} \times \text{g}^{-1} \times \text{mol}_{\text{Na}_2\text{SO}_4} \times (\text{mol}_{\text{Na}_2\text{SO}_4})^{-1} \times \text{mol}_{\text{Na}} \times (\text{mol}_{\text{Na}})^{-1}

=142.1×45.98−1×(19.42g−(NaNO3)g×0.2705) = 142.1 \times 45.98^{-1} \times (19.42 \text{g} - (\text{NaNO}_3)\text{g} \times 0.2705)

=3.090×(19.42g−(NaNO3)g×0.2705) = 3.090 \times (19.42 \text{g} - (\text{NaNO}_3)\text{g} \times 0.2705)

=60.01g−0.8358×(NaNO3)g = 60.01 \text{g} - 0.8358 \times (\text{NaNO}_3)\text{g}

Now we have a system of two equations with two unknowns.

(Na2SO4)g=79.94g−1.254×(NaNO3)g (\text{Na}_2\text{SO}_4)\text{g} = 79.94\text{g} - 1.254 \times (\text{NaNO}_3) \text{g}

(Na2SO4)g=60.01g−0.8358×(NaNO3)g (\text{Na}_2\text{SO}_4)\text{g} = 60.01 \text{g} - 0.8358 \times (\text{NaNO}_3)\text{g}

To solve, substitute the first equation into the second.

79.94g−1.254×(NaNO3)g=60.01g−0.8358×(NaNO3)g 79.94\text{g} - 1.254 \times (\text{NaNO}_3) \text{g} = 60.01 \text{g} - 0.8358 \times (\text{NaNO}_3)\text{g}

Then get NaNO3\text{NaNO}_3 on one side.

0.8358×(NaNO3)g−1.254×(NaNO3)g=60.01g−79.94g 0.8358 \times (\text{NaNO}_3)\text{g} - 1.254 \times (\text{NaNO}_3) \text{g} = 60.01 \text{g} - 79.94\text{g}

(NaNO3)g=(60.01g−79.94g)×(0.8358−1.254)−1 (\text{NaNO}_3) \text{g} = (60.01 \text{g} - 79.94\text{g}) \times (0.8358 - 1.254)^{-1}

Evaluate.

(NaNO3)g=−19.93g×−0.4182−1 (\text{NaNO}_3) \text{g} = −19.93\text{g} \times −0.4182^{-1}

(NaNO3)g=47.66g (\text{NaNO}_3) \text{g} = 47.66 \text{g}

Put back into the first equation and evaluate.

(Na2SO4)g=79.94g−1.254×47.66g (\text{Na}_2\text{SO}_4)\text{g} = 79.94\text{g} - 1.254 \times 47.66 \text{g}

(Na2SO4)g=79.94g−59.77g (\text{Na}_2\text{SO}_4)\text{g} = 79.94\text{g} - 59.77 \text{g}

(Na2SO4)g=20.17g (\text{Na}_2\text{SO}_4)\text{g} = 20.17 \text{g}

The percentages, by this method, are as follows.

  • NaCl=32.17%\text{NaCl} = 32.17\%
  • NaNO3=47.66%\text{NaNO}_3 = 47.66\%
  • Na2SO4=20.17%\text{Na}_2\text{SO}_4 = 20.17\%

Which is satisfying because it adds to exactly 100%.

32.17%+47.66%+20.17%=100% 32.17\% + 47.66\% + 20.17\% = 100\%

But it is different to the answers in the book.

  • NaCl=32.17%\text{NaCl} = 32.17\%
  • NaNO3=47.75%\text{NaNO}_3 = 47.75\%
  • Na2SO4=20.09%\text{Na}_2\text{SO}_4 = 20.09\%

Alternatively we can derive equations for NaNO3\text{NaNO}_3. First, from O\text{O}.

(NaNO3)g=(22.99+14.01+16.00×3)g1molNaNO3×(36.01g−(Na2SO4)g×16.00g×422.99g×2+32.07g+16.00g×4)×1molO16.00g×1molNaNO33molO (\text{NaNO}_3) \text{g} = \frac {(22.99 + 14.01 + 16.00 \times 3) \text{g}} {1 \text{mol}_{\text{NaNO}_3}} \times ( 36.01 \text{g} - (\text{Na}_2\text{SO}_4)\text{g} \times \frac {16.00 \text{g} \times 4} { 22.99 \text{g} \times 2 + 32.07 \text{g} + 16.00 \text{g} \times 4 } ) \times \frac{1 \text{mol}_{\text{O}}}{16.00 \text{g}} \times \frac{1 \text{mol}_{\text{NaNO}_3}}{3 \text{mol}_{\text{O}}}

=85.00g1molNaNO3×(36.01g−(Na2SO4)g×16.00g×4142.1g)×1molO16.00g×1molNaNO33molO = \frac {85.00 \text{g}} {1 \text{mol}_{\text{NaNO}_3}} \times ( 36.01 \text{g} - (\text{Na}_2\text{SO}_4)\text{g} \times \frac{16.00 \text{g} \times 4}{142.1 \text{g}} ) \times \frac{1 \text{mol}_{\text{O}}}{16.00 \text{g}} \times \frac{1 \text{mol}_{\text{NaNO}_3}}{3 \text{mol}_{\text{O}}}

=85.00g1molNaNO3×(36.01g−(Na2SO4)g×0.4504)×1molO16.00g×1molNaNO33molO = \frac {85.00 \text{g}}{1 \text{mol}_{\text{NaNO}_3}} \times(36.01 \text{g} - (\text{Na}_2\text{SO}_4)\text{g} \times 0.4504) \times \frac{1 \text{mol}_{\text{O}}}{16.00 \text{g}} \times \frac{1 \text{mol}_{\text{NaNO}_3}}{3 \text{mol}_{\text{O}}}

=85.00×g×(molNaNO3)−1×(36.01×g−(Na2SO4)×g×0.4504)×molO×16.00−1×g−1×molNaNO3×3−1×(molO)−1 = 85.00 \times \text{g} \times (\text{mol}_{\text{NaNO}_3})^{-1} \times( 36.01 \times \text{g} - (\text{Na}_2\text{SO}_4) \times \text{g} \times 0.4504 ) \times \text{mol}_{\text{O}} \times 16.00^{-1} \times \text{g}^{-1} \times \text{mol}_{\text{NaNO}_3} \times 3^{-1} \times (\text{mol}_{\text{O}})^{-1}

=85.00×16.00−1×3−1×(36.01×g−(Na2SO4)×g×0.4504)×g×g−1×molNaNO3×(molNaNO3)−1×molO×(molO)−1 = 85.00 \times 16.00^{-1} \times 3^{-1} \times( 36.01 \times \text{g} - (\text{Na}_2\text{SO}_4) \times \text{g} \times 0.4504 ) \times \text{g} \times \text{g}^{-1} \times \text{mol}_{\text{NaNO}_3} \times (\text{mol}_{\text{NaNO}_3})^{-1} \times \text{mol}_{\text{O}} \times (\text{mol}_{\text{O}})^{-1}

=85.00×16.00−1×3−1×(36.01×g−(Na2SO4)×g×0.4504) = 85.00 \times 16.00^{-1} \times 3^{-1} \times ( 36.01 \times \text{g} - (\text{Na}_2\text{SO}_4) \times \text{g} \times 0.4504 )

=85.00×48−1×(36.01×g−(Na2SO4)×g×0.4504) = 85.00 \times 48^{-1} \times ( 36.01 \times \text{g} - (\text{Na}_2\text{SO}_4) \times \text{g} \times 0.4504 )

=1.771×(36.01×g−(Na2SO4)×g×0.4504) = 1.771 \times ( 36.01 \times \text{g} - (\text{Na}_2\text{SO}_4) \times \text{g} \times 0.4504 )

=1.771×36.01×g−1.771×(Na2SO4)×g×0.4504 = 1.771 \times 36.01 \times \text{g} - 1.771 \times (\text{Na}_2\text{SO}_4) \times \text{g} \times 0.4504

=63.77g−0.7977×(Na2SO4)g = 63.77 \text{g} - 0.7977 \times (\text{Na}_2\text{SO}_4) \text{g}

Second, from the remaining Na\text{Na}.

(NaNO3)g=85.00g1molNaNO3×(19.42g−(Na2SO4)g×22.99g×2142.1g)×1molNa22.99g×1molNaNO31molNa (\text{NaNO}_3)\text{g} = \frac{85.00 \text{g}}{1 \text{mol}_{\text{NaNO}_3}} \times ( 19.42 \text{g} - (\text{Na}_2\text{SO}_4)\text{g} \times \frac{22.99 \text{g} \times 2}{142.1 \text{g}} ) \times \frac{1 \text{mol}_{\text{Na}}}{22.99 \text{g}} \times \frac {1 \text{mol}_{\text{NaNO}_3}} {1 \text{mol}_{\text{Na}}}

=85.00g1molNaNO3×(19.42g−(Na2SO4)g×0.3236)×1molNa22.99g×1molNaNO31molNa = \frac{85.00 \text{g}}{1 \text{mol}_{\text{NaNO}_3}} \times ( 19.42 \text{g} - (\text{Na}_2\text{SO}_4)\text{g} \times 0.3236 ) \times \frac{1 \text{mol}_{\text{Na}}}{22.99 \text{g}} \times \frac{1 \text{mol}_{\text{NaNO}_3}}{1 \text{mol}_{\text{Na}}}

=85.00×g×(molNaNO3)−1×(19.42g−(Na2SO4)g×0.3236)×molNa×22.99−1×g−1×molNaNO3×(molNa)−1 = 85.00 \times \text{g} \times (\text{mol}_{\text{NaNO}_3})^{-1} \times ( 19.42 \text{g} - (\text{Na}_2\text{SO}_4)\text{g} \times 0.3236 ) \times \text{mol}_{\text{Na}} \times 22.99^{-1} \times \text{g}^{-1} \times \text{mol}_{\text{NaNO}_3} \times (\text{mol}_{\text{Na}})^{-1}

=85.00×22.99−1×(19.42g−(Na2SO4)g×0.3236)×g×g−1×molNaNO3×(molNaNO3)−1×molNa×(molNa)−1 = 85.00 \times 22.99^{-1} \times ( 19.42 \text{g} - (\text{Na}_2\text{SO}_4)\text{g} \times 0.3236 ) \times \text{g} \times \text{g}^{-1} \times \text{mol}_{\text{NaNO}_3} \times (\text{mol}_{\text{NaNO}_3})^{-1} \times \text{mol}_{\text{Na}} \times (\text{mol}_{\text{Na}})^{-1}

=85.00×22.99−1×(19.42g−(Na2SO4)g×0.3236) = 85.00 \times 22.99^{-1} \times ( 19.42 \text{g} - (\text{Na}_2\text{SO}_4)\text{g} \times 0.3236 )

=3.697×(19.42g−(Na2SO4)g×0.3236) = 3.697 \times ( 19.42 \text{g} - (\text{Na}_2\text{SO}_4)\text{g} \times 0.3236 )

=3.697×19.42g−3.697×0.3236×(Na2SO4)g = 3.697 \times 19.42 \text{g} - 3.697 \times 0.3236 \times (\text{Na}_2\text{SO}_4)\text{g}

=71.80g−1.196×(Na2SO4)g = 71.80 \text{g} - 1.196 \times (\text{Na}_2\text{SO}_4)\text{g}

Now we have a different system of equations.

(NaNO3)g=63.77g−0.7977×(Na2SO4)g (\text{NaNO}_3)\text{g} = 63.77 \text{g} - 0.7977 \times (\text{Na}_2\text{SO}_4) \text{g}

(NaNO3)g=71.80g−1.196×(Na2SO4)g (\text{NaNO}_3)\text{g} = 71.80 \text{g} - 1.196 \times (\text{Na}_2\text{SO}_4)\text{g}

To solve.

63.77g−0.7977×(Na2SO4)g=71.80g−1.196×(Na2SO4)g 63.77 \text{g} - 0.7977 \times (\text{Na}_2\text{SO}_4) \text{g} = 71.80 \text{g} - 1.196 \times (\text{Na}_2\text{SO}_4)\text{g}

(Na2SO4)g=(71.80g−63.77g)×(1.196−0.7977)−1 (\text{Na}_2\text{SO}_4)\text{g} = (71.80 \text{g} - 63.77 \text{g}) \times (1.196 - 0.7977)^{-1}

(Na2SO4)g=8.030g×0.3983−1 (\text{Na}_2\text{SO}_4)\text{g} = 8.030 \text{g} \times 0.3983^{-1}

(Na2SO4)g=20.16g (\text{Na}_2\text{SO}_4)\text{g} = 20.16 \text{g}

(NaNO3)g=63.77g−0.7977×20.16g (\text{NaNO}_3)\text{g} = 63.77 \text{g} - 0.7977 \times 20.16 \text{g}

(NaNO3)g=63.77g−16.08g (\text{NaNO}_3)\text{g} = 63.77 \text{g} - 16.08 \text{g}

(NaNO3)g=47.69g (\text{NaNO}_3)\text{g} = 47.69 \text{g}

Yielding this set of percentages.

  • NaCl=32.17%\text{NaCl} = 32.17\%
  • NaNO3=47.69%\text{NaNO}_3 = 47.69\%
  • Na2SO4=20.16%\text{Na}_2\text{SO}_4 = 20.16\%

Mixing the two sets, so we have (Na2SO4)g(\text{Na}_2\text{SO}_4)\text{g} derived from the remaining Na\text{Na} and (NaNO3)g(\text{NaNO}_3)\text{g} from the O\text{O}, we can get this system.

(Na2SO4)g=60.01g−0.8358×(NaNO3)g (\text{Na}_2\text{SO}_4)\text{g} = 60.01 \text{g} - 0.8358 \times (\text{NaNO}_3)\text{g}

(NaNO3)g=63.77g−0.7977×(Na2SO4)g (\text{NaNO}_3)\text{g} = 63.77 \text{g} - 0.7977 \times (\text{Na}_2\text{SO}_4) \text{g}

With this solution.

(NaNO3)g=63.77g−0.7977×(60.01g−0.8358×(NaNO3)g) (\text{NaNO}_3)\text{g} = 63.77 \text{g} - 0.7977 \times (60.01 \text{g} - 0.8358 \times (\text{NaNO}_3)\text{g})

(NaNO3)g=63.77g+(−0.7977×60.01g+−0.7977×−0.8358×(NaNO3)g) (\text{NaNO}_3)\text{g} = 63.77 \text{g} + ( -0.7977 \times 60.01 \text{g} + -0.7977 \times -0.8358 \times (\text{NaNO}_3)\text{g} )

(NaNO3)g=63.77g+−47.869977g+0.6667×(NaNO3)g (\text{NaNO}_3)\text{g} = 63.77 \text{g} + -47.869977 \text{g} + 0.6667 \times (\text{NaNO}_3)\text{g}

1×(NaNO3)g+−0.6667×(NaNO3)g=63.77g+−47.87g 1 \times (\text{NaNO}_3)\text{g} + -0.6667 \times (\text{NaNO}_3)\text{g} = 63.77 \text{g} + -47.87 \text{g}

(1+−0.6667)×(NaNO3)g=63.77g+−47.87g (1 + -0.6667) \times (\text{NaNO}_3)\text{g} = 63.77 \text{g} + -47.87 \text{g}

(NaNO3)g=63.77g+−47.87g×(1+−0.6667)−1 (\text{NaNO}_3)\text{g} = 63.77 \text{g} + -47.87 \text{g} \times (1 + -0.6667)^{-1}

(NaNO3)g=15.9g×0.3333−1 (\text{NaNO}_3)\text{g} = 15.9 \text{g} \times 0.3333^{-1}

(NaNO3)g=47.70g (\text{NaNO}_3)\text{g} = 47.70 \text{g}

(Na2SO4)g=60.01g−0.8358×47.70g (\text{Na}_2\text{SO}_4)\text{g} = 60.01 \text{g} - 0.8358 \times 47.70 \text{g}

(Na2SO4)g=20.14g (\text{Na}_2\text{SO}_4)\text{g} = 20.14 \text{g}

Yielding this set of percentages.

  • NaCl=32.17%\text{NaCl} = 32.17\%
  • NaNO3=47.70%\text{NaNO}_3 = 47.70\%
  • Na2SO4=20.14%\text{Na}_2\text{SO}_4 = 20.14\%