Unit  3  Area  of  Study  1:  Chemical  Analysis

Gravimetric  Analysis

Go to the lesson for this slide deck: https://edrolo.com.au/vce/subjects/chemistry/vce-chemistry/aos-1-chemical-analysis/chemical-analysis-gravimetric-analysis/outcome-1/#watch

Outcome  1    

On  completion  of  this  unit  the  student  should  be  able  to  evaluate  the  suitability  of  techniques  and  instruments  used  in  chemical  analyses.    Key  Knowledge  � Gravimetric  analysis  � the  application  of  chemical  equations  to  (volumetric  and)  gravimetric  analyses  

Gravimetric  analysis  •  This  form  of  quantitative  analysis  involves  Find  the  mass  of  a  part  of  the  sample  and  relating  it  back  to  the  sample,  often  as  a  percentage  by  mass  

•  The  part  of  the  sample  we  are  interested  in  may  be  the  water  in  the  product,  or  perhaps  we  might  be  focussing  on  a  particular  ion  that  is  present.  

Measuring  water  content  •  Many  solids  have  a  high  proportion  of  water.  •  Dehydration  removes  the  water  and  decreases  the  mass  

•  This  may  be  heating  at  just  above  100  °C  •  Mass  loss  =  mass  of  water  originally  present  

•  %  water  content  =  m(water)  ×  100                                                                                                                                                                m(sample)  

Weigh  the  sample  

Heat  the  sample  in  an  oven  just  above  100  °C  

Allow  the  sample  to  cool  in  a  desiccator  

Reweigh  the  sample  

This cycle is repeated until the sample assumes a constant mass

Finding  the  composition  of  a  mixture  

1.  Dissolve  the  mixture  in  water  2.  Add  an  ionic  solution  which  will  

form  a  precipitate  with  the  ion  that  is  required  (eg  SO42-­‐).  

3.  This  new  solid  is  then  Filtered  out,  dried  and  weighed.

The  most  common  precipitates  

Element  you  wish  to  analyse  

Substance  that  is  added  to  create  precipitate  

Formula  of  precipitate  

Chlorine     Silver  nitrate   AgCl  

Bromine   Silver  nitrate   AgBr  

Sulfur   Barium  chloride   BaSO4  

Application  of  gravimetric  analysis  �  Gravimetric  analysis  is  used  in  the  food  industry  to  determine  the  amount  of  salt  in  products.  

� When  an  AgNO3  solution  is  added              to  the  dissolved  food,  chloride  ions              (present  as  NaCl)  are  precipitated  out        as  AgCl.      

Calculations  Find  n(AgCl)  using:  n(AgCl)  =    Then:  

n(AgCl)  =  n(Cl-­‐)  =  n(NaCl)  � Convert  n(NaCl)  to  m(NaCl)  using                m(NaCl)  =  n(NaCl)  ×  M(NaCl)    � %(NaCl)  =  m(NaCl)  ×  100                    m(food)          1  

m(AgCl)M(AgCl)

Experiment      Analysis  of  Lawn  food  

Lawn  food  contains  many  nutrients  required  by  plants  such  as  sulfate  ions,  SO42-­‐,  nitrate  ions,  NO3-­‐  and  phosphate  ions,  PO43-­‐.    Using  gravimetric  analysis  to  ;ind  the  amount  of  sulfate  ions  in  a  sample  of  lawn  food  1.  Dissolve  lawn  food  in  water  2.  Add  barium  chloride  solution  3.  Sulfate  ions  precipitate  out  as  barium  

sulfate                      BaCl2(aq)  +  SO42-­‐  (aq)  →  BaSO4(s)  +  2Cl-­‐(aq)  

4.  Filter,  wash  and  dry  the  BaSO4  to  obtain  a  pure  sample  of  precipitate.  

 6.  During  washing  stages  Filtrate  is  tested  for  

presence  of  left  over  chloride  ions  by  adding  silver  nitrate.    AgNO3(aq)+  Cl-­‐  →  AgCl(s)  +  NO3-­‐(aq)  

 •   If  ions  are  in  the  Filtrate,  then  they  have  

probably  also  stuck  to  the  solid  and  its  mass  will  be  too  high.      

 

6.  Barium  sulfate  is  dried,  then  weighed.    8.  Using  M(BaSO4),  n(BaSO4)    is  calculated  so  

n(SO42-­‐)  can  be  deduced.      n(SO42-­‐)  =  n(BaSO4)  

8.  m(SO42-­‐)  =  n(SO42-­‐)  ×  M(SO42-­‐)    9.  m(S)  =  n(SO42-­‐)  ×  M(S)    

10. %  S  =    m(S) 100

m(sample) 1×

Sample  exam  question  An  8.64g    sample  of  rock  was  thought  to  be  made  up  of  CaCO3  which  reacted  with  HCl  according  to:  2HCl(aq)  +  CaCO3(s)  →  CaCl2(aq)    +    H2O(l)  +  CO2(g)  and  1.55g  of  insoluble  SiO2.  (a) Calculate  the  expected  percentage  of  CaCO3  in  the  rock  sample  (b) The  dissolved  calcium  ions  were  precipitated  out  as  CaC2O4.H2O  which  was  collected,  washed  and  dried.  It  was  converted  to  CaO  (M  =  56.1gmol-­‐1)  and  3.87g  was  obtained.  Calculate  the  %  CaCO3  in  the  rock  sample.  (signiFicant  Figures)  

(VCAA  June  2007  Q3)  

(a) m(CaCO3)  =  8.64  –  1.55  =  7.09g              %(CaCO3)  =      b)  n(CaO)  =            mol              n(Ca2+)  =  0.06898  mol  =  n(CaCO3)    m(CaCO3)  =  0.06898  ×  100.1  =  6.905g    %(CaCO3)  =                                            (3  sig  Figs)  

   

7.09 100 82.1%8.64 1

× =

3.87 0.0689856.1

=

6.905 100 79.9%8.64 1

× =

Answers  

Xylose is a compound that has five carbon atoms in each molecule and contains 40% carbon by mass. What is the molar mass of xylose?

A. 30

B. 67

C. 150

D. It cannot be determined without further information.

(VCAA Q3 June 2008)

QUESTION 1

Xylose is a compound that has five carbon atoms in each molecule and contains 40% carbon by mass. What is the molar mass of xylose? Solution M(C) = 12.0 gmol-1 1 mol xylose molecules contains 5 mol C atoms (5 C atoms per molecule, so 5 mol of C atoms per 1 mol of molecules) m(5 mol C atoms) = 60.0 g 40% of xylose is carbon, so 40% of M(xylose) = 60.0 This can also be written as 0.40 × M(xylose) = 60.0 M(xylose) = = 150.0 gmol-1

Alternative C is the correct answer

A.  30 B.  67 C.  150 D.  It cannot be determined without further information.

QUESTION 1 (ANSWER)

60.00.40

The amount of calcium carbonate (:molar mass = 100.1 g ) in the ore dolomite can be determined by gravimetric analysis. The dolomite sample is dissolved in acid and the calcium ions () present are precipitated as calcium oxalate ( :molar mass = 128.1 g ). The calcium oxalate is foltered, dried and strongly heated to form calcium oxide (CaO); molar mass = 56.1 g ). In one analysis the mass of dolomite used was 3.72 g. The mass of calcium oxide formed was found to be 1.24 g. The percentage of calcium carbonate in the dolomite sample is closest to

A: 26.

B: 33.3

C: 56.0

D: 59.5 (VCAA Q8 June 2005)

QUESTION 2

The amount of calcium carbonate (:molar mass = 100.1 g ) in the ore dolomite can be determined by gravimetric analysis. The dolomite sample is dissolved in acid and the calcium ions () present are precipitated as calcium oxalate ( :molar mass = 128.1 g ). The calcium oxalate is foltered, dried and strongly heated to form calcium oxide (CaO); molar mass = 56.1 g ). In one analysis the mass of dolomite used was 3.72 g. The mass of calcium oxide formed was found to be 1.24 g. The percentage of calcium carbonate in the dolomite sample is closest to Solution n(CaO) = n(Ca2+) = n(CaO) = n(CaCO3) = 0.0221 m(CaCO3) in dolomite = 0.0221 × 100.1 = 2.21g %(CaCO3) = Alternative D is the correct answer

a)  26.9 b)  33.3 c)  56.0 d)  59.5

QUESTION 2 (ANSWER)

2.21 100 59.5%3.72 1

× =

2.21 100 59.5%3.72 1

× =

When 2.54 g of solid iodine reacts with excess chlorine and the unreacted chlorine is evaporated, 4.67 g of a yellow product remains. The empirical formula of the product is

A: ICl2

B: ICl3

C: ICl4

D: ICl5

(VCAA Q4 June 2007)

QUESTION 3

When 2.54 g of solid iodine reacts with excess chlorine and the unreacted chlorine is evaporated, 4.67 g of a yellow product remains. The empirical formula o the product is Solution The yellow product is only made up of chlorine and iodine In 4.67g of the yellow product, 2.54g is iodine and (4.67-2.54) 2.13g is chlorine Empirical formula: I : Cl 2.54g : 2.13g 0.0200 : 0.0600

1 : 3

The empirical formula is ICl3 so the correct answer is B A: ICl2 B: ICl3 C: ICl4 D: ICl5

QUESTION 3 (ANSWER)

2.54 2.13 : 126.9 35.5