Question

Sam records the mass of his evaporating dish as 6.251 g.


He records the mass of the evaporating dish and the sample of hydrate as 16.864 g.


After heating the sample in the evaporating dish to constant weight, the mass of them combined is 11.13 g.


How many moles of water were removed from the sample by the heating process?


Report your answer with three digits after the decimal.

Answer 1

Answer:

• 0.318 moles

Explanation:

1. Mass of water removed

• Mass of water removed = mass of the evaporating dish and the sample of hydrate - mass of the evaporating dish and the sample after heating

• Mass of water evaporated =  16.864 g - 11.13 g = 5.734 g

This amount must be rounded to two decimals, because the least precise measure (11.13 g) contains two decimals.

• Mass of water evaporated = 5.73 g

2. Moles of water removed from the sample by the heating process

You must use the molar mass of water to convert 5.73 g of water to number of moles.

• Molar mass of water: 18.015 g/mol

• Number of moles = mass in grams / molar mass = 5.73 g / 18.015 g/mol = 0.318 moles.

Answer 2

Based on the formula for calculating moles of a substance, the moles of water removed is 0.318 moles.

What is the mass of water removed?

The mass of water removed is calculated thus:

• Mass of water removed = mass of the evaporating dish and sample of hydrate - mass of the evaporating dish and the sample after heating

Mass of water evaporated =  16.864 g - 11.13 g

Mass of water evaporated = 5.734 g

What is the moles of water removed from the sample by the heating process?

The moles of water removed is calculated using the formula:

• Moles = mass/ molar mass

Molar mass of water: 18.015 g/mol

Number of moles = 5.734 g / 18.015 g/mol

Moles of water = 0.318 moles.

Therefore, the moles of water removed is 0.318 moles.

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