Chemical Formulas Questions efore beginning this experiment in the laboratory, y

Question

Chemical Formulas Questions efore beginning this experiment in the laboratory, you should be able to answer the following questions. I. Give the chemical sy 3. Define the term compound. 4. The molecular formula for benzene is CHe. What is its empirical formula? 5. Balance the following equation: Al(s)+Citg)- AICH(s). 6. What is the percentage composition of PbCO,? A substance was found by analysis to contain 45.57% tin and 54.43% chlorine, what is the empirical formula for the substance? 7. 63 Copyright 2015 Pearson Education, Inc.

Explanation / Answer

1.Lead - Pb Iron - Fe Fluorine - F Potassium - K
2. KBr = 39 + 79 =118 PbCO3 = 207+12+3x16 = 267

3.compound is a substance that results from a combination of two or more different chemical element s, in such a way that the atom s of the different elements are held together by chemical bonds.

4. CH is the empirical formula of Benzene

5. 2 Al + 3 Cl2 = 2AlCl3

6. Molecular weight of PbCO3 = 267

% composition of lead = 207/267 x 100 =77.52 %

% composition of Carbon = 12/267 x 100 = 4.494%

% composition of Oxygen = 48/207 = 23.188 %

7. Assume 100g of compound

Tin = 45.57g Chlorine = 54.43g

moles of tin = 45.57/118 = 0.386

moles of chlorine = 54.43/35 = 1.555

Now divide the number of moles of tin with chlorine

1.555/0.386 = 4

The empirical formula is TiCl4

8.the law of definite composition, states that a chemical compound always contains exactly the same proportion of elements by mass.

9. Empirical formula give the simplest ratio of elements in a compound.

Molecular formula give the number of atoms of elements in the molecular compound.

10. Molecular wt of CaI2 = 293 g/mol

No of moles = 1.73

Mass = 1.73 x 293 506.89

11. 2 mol decompose to give 3 moles of N2

So 1.6 mol gives3/2 x 1.6 = 2.4 mol

b) 7 g of N2 means 0.25 moles

No of moles of NaN3 = 2x0.25/3 = 0.166 moles

0.166 x 28 = 4.64 g To get 7 g of Nitrogen, 4.64 g of NaN3 is required

12. % compostion of elements : C = 3.758/5.325x100 = 70.572%

Hydrogen = 0.316/5.325 = 5.934%

Oxygen = 1.251/5.325 =23.492%

Consider 100 g of compound

Moles of C =70.572/12 = 5.881 ; moles of oxygen = 23.492/16= 1.468 ; moles of H=5.934/1 = 5.9

Divide the moles by least number of moles that is 1.468

C;5.881/1.468 = 4 ; Oxygen =1 ; Hydrogen = 5.9/1.468 = 4 (rounded off to the neares whole numbers)

Empirical formula = C4H4O

empirical weight = 12x4+4x1+16 = 68

molar mass = 130 ; 130/68 approx equal to 2

So the molecular formula is C8H8O2

13. Molecular wt = 92.02

Consider 100 g of compound

moles of oxygen = 69.57/16 = 4.348 ; moles of nitrogen = 30.43/14=2.173

Dividing both by 2.173

N=1 ; O = aprrox 2

Empirical formula is NO2

Molecular wt of NO2 = 14+16x2 = 46 ; 92/46=2.0

Formula of compound = N2O4

2NO + O2 2NO2

2NO2 N2O4

14. Moles of sodium = 0.131g/23 = 0.0056

Every mol contains 6.022 x 10^23 atoms

Number of atoms in 0.0056 mol of sodium = 0.0056 x 6.022 x 10^23 = 3.4299 x 10^21 atoms

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