Chapter 27: Q37P (page 769)

Determination of sulfur by combustion analysis produces a mixture of ${SO}_{2}$ and ${SO}_{3}$ that can be passed through $H_{2} O_{2}$ to convert both into $H_{2} {SO}_{4}$ , which is titrated with standard base. When 6.123m g of a substance were burned, the $H_{2} {SO}_{4}$ required 3.01mL of0.010576MNaOHfor titration. Find wt% sulfur in the sample.

Short Answer

Expert verified

The weight percent of sulfur in the sample is 0.124 .

Step by step solution

Derfining the weight percent.

The Weight Percentage is simply the ratio of a solute's mass to the mass of a solution multiplied by 100. The Weight Percentageis also referred to as the Mass Percentage.

$Percentbyweight = \frac{gram of solute}{100 g of solution}$ $Percentbyweight = \frac{gram of solute}{100 g of solution}$

Determining the moles of NaOH and H2So$

Here of were required for titration of when of a substance were burned. We need to find the weight percent of sulfur in the sample.

Consider the following reaction:

$H_{2} {SO}_{4} + 2 NaOH \to 2 H_{2} O + {Na}_{2} {SO}_{4} n (NaOH) = c \times V n (NaOH) = 0.01576 M \times 3.01 \times 10^{- 3} L n (NaOH) = 4.744 \times 10^{- 5} mol n (H_{2} {SO}_{4}) = 2.372 \times 10^{- 5} mol$

Calculating the weight percent.

Then we will find the weight percent of sulfur in the sample:

$wt % = \frac{m (S)}{m (sample)} wt % = \frac{0.7607 mg}{6.123 mg} wt % = 0.124$

Therefore the weight percent of sulfur in the sample is 0.124 .

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Short Answer

Step by step solution

Derfining the weight percent.

Determining the moles of NaOH and H2So$

Calculating the weight percent.

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