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# 1.A 0.5538 g sample of a pure soluble bromide compound is dissolved in water, and all...

1.A 0.5538 g sample of a pure soluble bromide compound is dissolved in water, and all of the bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate. The mass of the resulting AgBr is found to be 1.1298 g. What is the mass percentage of bromine in the original compound? %

2. A student determines the manganese(II) content of a solution by first precipitating it as manganese(II) hydroxide, and then decomposing the hydroxide to manganese(II) oxide by heating. How many grams of manganese(II) oxide should the student obtain if his solution contains 37.0 mL of 0.419 M manganese(II) nitrate? g

3. A 13.36 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 19.58 grams of CO2 and 8.017 grams of H2O are produced. In a separate experiment, the molar mass is found to be 60.05 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

4. A 3.605 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 5.374 grams of CO2 and 1.650 grams of H2O are produced. In a separate experiment, the molar mass is found to be 118.1 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

1)

1st find the mass of Br in AgBr

Molar mass of AgBr,

MM = 1*MM(Ag) + 1*MM(Br)

= 1*107.9 + 1*79.9

= 187.8 g/mol

mass(AgBr)= 1.1298 g

use:

number of mol of AgBr,

n = mass of AgBr/molar mass of AgBr

=(1.13 g)/(1.878*10^2 g/mol)

= 6.016*10^-3 mol

This is number of moles of AgBr

one mole of AgBr contains 1 moles of Br

use:

number of moles of Br = 1 * number of moles of AgBr

= 1 * 6.016*10^-3

= 6.016*10^-3

Molar mass of Br = 79.9 g/mol

use:

mass of Br,

m = number of mol * molar mass

= 6.016*10^-3 mol * 79.9 g/mol

= 0.4807 g

same will be the mass of Br in sample

mass % of Br = mass of Br * 100 / mass of sample

= 0.4807 * 100 / 0.5538

= 86.8 %

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