(a) How many moles of potassium ions are in 306 g alum? - mol (b) What is the ma

Question

(a) How many moles of potassium ions are in 306 g alum?
-   mol
(b) What is the mass % S in alum (to 3 significant figures)?
-   %
(c) What is the maximum amount of alum (in metric tons) that can be prepared from 18.2 metric tons of scrap aluminum (1 metric ton = 1000 kg)?
- metric tons

2) The bitter-tasting compound quinine is a component of tonic water and is used as a protection against malaria. It contains only C, H, N and O. When a sample of mass 0.487 g was burned, 1.321 g of carbon dioxide, 0.325 g of water, and 0.0421 g of nitrogen were produced. The molar mass of quinine is 324 g/mol. Determine the empirical and molecular formulas of quinine. (Type your answer using the format CO2 for CO2 and use the order CHNO)
- empirical
  
-molecular

3) Calculate each of the following quantities.

Explanation / Answer

1) Alum formula : KAl(SO4)2.12.H2O

(a) 1 mole of Alum has 1 K+ ion

moles of alum = 306/474.3884 = 0.645 mols

So, moles of K+ ions = 0.645 mols

(b) 1 mole of alum has 2 mols of S in it

mass of S (%) in alum = (2 x 32.065/474.3884) x 100 = 13.518 %

(c) when we start with 18.2 metric ton = 18.2 x 1000 x 1000 g of Al

moles of Al = 18.2 x 10^6/26.982 = 6.75 x 10^5 mols

amount of alum formed = 6.75 x 10^5 x 474.3884 = 3.20 x 10^8 g = 320.212 metric ton

2) Find C, H and N in the sample in grams

C = 1.321 x 12.011/44.0098 = 0.361 g

H = 0.325 x 2.0158/18.0152 = 0.0364 g

N = 0.0421 g

O = 0.487 - 0.36 + 0.036 + 0.0421 = 0.0475 g

moles of C = 0.361/12.011 = 0.03

moles of H = 0.036/1.008 =0.036

moles of N = 0.0421/14 = 0.003

moles of O = 0.0489/16 = 0.003

divide by smallest factor,

C = 0.03/0.0003 = 10

H = 0.036/0.003 = 12

N = 0.003/0.003 = 1

O = 0.003/0.003 = 1

So empirical formula becomes = C10H12NO

empirical formula mass = 12 x 10 + 1 x 12 + 14 + 16 = 162 g

molar mass of quinine/emirical formula mass = 324/162 = 2

So molecular formula of quinine = C20H24N2O2

3) (a) 3.9 x 10^20 molecules of NO2

= (3.9 x 10^20/6.022 x 120^23) x 46.0055

= 0.0298 g = 2.98 x 10^-5 kg of NO2

(b) moles of Cl atoms in 0.0379 g C2H4Cl2

1 mole of C2H4Cl2 has 2 moles of Cl

98.96 g of C2H4Cl2 has 32 g of Cl

So, 0.0379 g of C2H4Cl2 will have = 0.0379 x 32/98.96 = 0.0123 g of Cl

moles of Cl = 0.0123/16 = 7.69 x 10^-4 moles

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