1. A student determined percent of cr in a sample of an unknown chloride gravime

Question

1. A student determined percent of cr in a sample of an unknown chloride gravimetrically by He dissolving 0.907 g of solid unknown in water, adding excess of AgNO3 and precipitating Ag obtained 2.350 g of AgCI. Using these data, calculate chloride in the unknown. (Give set-up and numerical answer. If you use more than one step in your calculations, show all work.) 2. The only other component of the unknown mentioned in #1 is known to be Calcium. a) What is the per cent by weight of Calcium in the unknown? b) Using your answers to questions #1 and 2a, determine the empirical formula of the unknown chloride. SHOW ALL WORK!

Explanation / Answer

(1)

Given:

Mass of Unknown = 0.907 g

Mass of AgCl preciptated = 2.35 g

Data:

Molar mass of Cl- = 35.5 g/mol

Molar mass of AgCl = 143.32 g/mol

Solution:

No. of moles of AgCl = mass of AgCl / Molar mass of AgCl = 2.35 g / 143.32 g/mol = 0.0164 moles

there will be 1 mole of Cl- present in each mole of AgCl

so No of moles of Cl- = 0.0164 moles

Hence mass of Cl- dissolved = No. of moles of Cl- * Molar mass of Cl- = 0.0164 moles * 35.5 g/mol = 0.582 g

mass % Cl- = Mass of Cl- in unknown / Mass of unknown = 0.582 g / 0.907 g = 0.6418

mass % Cl- = 64.18 % Answer

2)

Mass of Calcium in unknown = mass of Unknown - mass of Cl- = 0.907 - 0.582 = 0.325 g

mass % calcium = Mass of calcium in unknown / Mass of unknown = 0.325 g / 0.907 g = 0.3582

mass % Calcium = 35.82 % Answer

(b) molar mass of calcium = 40 g/mol

No. of moles of Ca = Mass of calcium / molar mass of calcium = 0.325 / 40 = 0.008125 moles

No. of moles of Cl = 0.0164 moles

dividing both by 0.008125

we will get Ca:Cl = 1:2

Emprical formula is CaCl2

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