CHEMICAL EXPERIMENTS AND THE MOL

I. What is the percent by mass of C  and O in 500 g of CO₂?

First get the mass of 1  mol….  12.0 + 2(16.0) = 44.0 g

So % C =  12.0 / 44.0  =   27.3 %

And O  =   32.0/ 44.0  =   72.7 %   or 100 – 27.3

In a  500 g sample, there is 500 x .273  = _______carbon

II. If you know the %mass then you can calculate the

Simplest Formula or Empirical Formula

When 100 g of  H_xO_y à H     +   O

                                                                                    11.1g      88.9g

The ratio of   mass of O / mass of H =  88.9 g/ 11.1g =  8

doesn’t tell us anything about the formula…. only that Oxygen by weight makes up 8 times the mass of Hydrogen.

mol of H  = 11.1g  ( 1mol/1.0g) = 11.1 mol H

mol of O =  88.9g ( 1mol/16.0g)=  5.56 mol H

 The ratio of H/O =  11.1/5.56=  2.00

 does tell us the simplest or empirical formula  must be H₂O .

Most ratios come out close to a whole number.(experimental error)

What other ratios will tell you different formula?

1.5  =   3/2                   A₃B₂

2.5 =  5/2                     A₅B₂

3.5  =  7/2        A₇B₂

1.3333  = 4/3

Try this one….

but the real formula (molecular formula) could be

                                                                          H₂O₁

  H₄O₂

                                                                          H₆O₃

                                                                          H₈O₄ 

↓

We need to know the molecular weight of the unknown  compound to determine its true MOLECULAR FORMULA:

What if the molecular weight was 54 g/mol.

Which one would it be?

H₆O₃

Really it is 18.0 g/mol … so the unknown formula is H₂O  !

PRACTICE:

- Simplest Formula Worksheet

- Heath Text p. 123  Q 17- 27

Answers

17. CH

18. SO₃

19. P₂O₅

20. W₂C

21. C₆N₅O₆H₆

22. CH₂O

23. C₂H₅, C₄H₁₀

24. % Al = 36.08, % S= 63.92

25. FeS  Iron (II) sulphide

26. Cu₂S   Copper (I) sulphide

27. NiC₂N₂ or Ni(CN)₂  nickel (II) cyanide