Question

A sample of potassium phosphate octahydrate (K3PO4•8H2O) is heated until 7.93 grams of water are released. How many grams did the original hydrate weigh?

Answer 1

Answer: The mass of original hydrate is 19.63 grams.

Explanation:

We are given:

Mass of water released = 7.93 grams

We are given a chemical compound known as potassium phosphate octahydrate having chemical formula of $K_3PO_4.8H_2O$

Mass of $K_3PO_4.8H_2O$ = 356.4 grams

Mass of 8 water of crystallization = (8 × 18) = 144 grams

By applying unitary method, we get:

144 grams of water is released when 356.4 grams of salt is heated.

So, 7.93 grams of water will be released when = $\frac{356.4g}{144g}\times 7.93g=19.63g$ of salt is heated.

Hence, the mass of original hydrate is 19.63 grams.

Answer 2

The original sample of potassium phosphate octahydrate had a mass of 19.6 grams. When it was heated, it released 7.93 grams of water.

Further Explanation:

For every mole of the compound potassium phosphate octahydrate, there are 8 moles of water of hydration which can be removed from the crystal by heating without altering the chemical composition of the substance.

To determine how much original sample was used, the amount of water released upon heating may be used as well as the mole ratio of the water of hydration with the compound itself following the steps below:

1. Convert mass of water released to moles.
2. Use the mole ratio of water of hydration to the compound (8 mol water for every mol of potassium phosphate octahydrate) to get the moles of original sample.
3. Convert the moles of original sample to grams.

STEP 1: Convert 7.93 g water to moles.

$moles \ of\ H_{2}O \ = 7.93 \ g \ H_{2}O \ (\frac{1 \ mol \ H_{2}O}{18.00 \ g \ H_{2}O})\\\boxed {moles \ of \ H_{2}O \ = 0.4406 \ mol}$

STEP 2: Calculate the moles of original sample using the mole ratio: 1 mol K3PO4 8H2O : 8 mol H2O.

$moles \ of \ K_{3}PO_{4}\ 8H_{2}O \ = 0.4406 \ mol \ H_{2}O \ (\frac{1 \ mol \ K_{3}PO_{4}\ 8H_{2}O \ }{8 \ mol \ H_{2}O})\\\\\boxed {moles \ of \ K_{3}PO_{4}\ 8H_{2}O \ = 0.0551 \ mol}$

STEP 3: Convert the moles of original sample to mass.

$mass \ of \ K_{3}PO_{4}\ 8H_{2}O = 0.0551 \ mol \ K_{3}PO_{4}\ 8H_{2}O \ (\frac{356.3885 \ g}{1 \ mol\ K_{3}PO_{4}\ 8H_{2}O})\\ mass \ of \ K_{3}PO_{4}\ 8H_{2}O \ = 19.637 \ g$

Following the significant figures of the given, the final answer should be:

$\boxed {mass \ of \ K_{3}PO_{4}\ 8H_{2}O = 19.6 \ g}$

Learn More:

1. Learn more about water of hydration brainly.com/question/6053815
2. Learn more about mole conversion brainly.com/question/12979299
3. Learn more about percent hydrate brainly.com/question/12398621

Keywords: water of hydration, hydrate