59% LD 6:33 PM iPad sacct.csus.edu 1. A student obtains the following data: Mass

Question

59% LD 6:33 PM iPad sacct.csus.edu 1. A student obtains the following data: Mass of test tube: 27.2877 g 28.5298 g Mass of test tube and hydrate: Mass of test tube and anhydrous residue after heating 28.0501 g The laboratory instructor identifies the sample as strontium chloride hydrate. (SrCl2.nH20) a) Calculate the mass percent of water in the hydrate? Answer: b) Calculate the number of moles of water in the hydrate sample that were driven off by heating? Answer: c) Calculate the mole ratio of water to salt in the hydrate, round to the nearest whole number. ("n" in srcl2.nH20) Answer: d) Write the correct formula for the hydrate and give the correct name? Formula: Name: Page 1 of 6 Rev. F14

Explanation / Answer

1.

Mass of test tube = 27.2877 g

Mass of test tube + hydrate = 28.5298 g

mass of test tube + anhydrous = 28.0501 g

Mass of hydrate = 28.5298 -27.2877 = 1.2421 g

Mass of water lost = 28.5298 - 28.0501 = 0.4797 g

a)

That means, 1.2421 g of hydrate form has 0.4797 g of water

100 g of hydrate form contains (0.4797 / 1.2421 ) x 100 = 38.62 %

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b)

Mass of water = 0.4797 g

Moles of water = 0.4797 g / 18 g/mol = 0.02665 mol (molar mass of water = 18 g/mol)

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c)

Mass of water = 0.4797 g

Moles of water = 0.02665 mol

Mass of salt = 1.2421 g - 0.4797 g = 0.7624 g

(Molar mass of SrCl2 = 158.53 g/mol)

Moles of salt = 4.81 x 10^-3 mol

Ratio of water to salt = 0.02665 : 0.00481 = 6:1

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d)

SrCl2.6H2O

Strontium chloride hexahydrate

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