Adding, Subtracting, Multiplying and Dividing Significant Figures

I observed in my classes that my students are always asking if how many decimal places are they going to round off the result of their calculations.  As a student they must know how to handle significant figures in doing calculations in addition, subtraction, multiplication and division.  You may have correct answer but the way to round off your answer to proper number of significant figures also matters.

There are rules to follow in using significant figures in calculations.
1.  In addition and subtraction, the calculated result must be rounded off to proper number of digits following the number which has least number of digits after the decimal point.   Let us have an example

                                                 90.335   cm   (3 digits after the decimal point)

                                           +      2.1       cm   (1 digit after the decimal point)
                                           ______________
                                                92.435   cm
                                     or        92.4  cm is the final answer.

The result is rounded off to one digit after the decimal point following the rule that in rounding off, answer in addition and subtraction, it should follow the number which has least number of digits after the decimal point.

Now in rounding off numbers we have to consider the number to be dropped so that we can decide if the preceding digit can be retained or should be increased by 1.  If the number to be dropped is 5 and above, the preceding digit should be increased by 1 but if the number to be dropped is less that 5 the preceding digit should be retained.  In the above example, since there should be 1 digit after the decimal point, the number to be dropped is 3 which is less than 5 and so we retain the value of the preceding digit which is 4.

2.  In multiplication and division, the calculated result should be rounded off to proper number of significant figures following the value or number which has least number of significant figures.  Let us have an example:

2.8  x  4.5039  =  12.61092    or 13

The above example was rounded off to 13 having only two significant figures following the least number of significant figures from the given,  two significant figures for 2.8 and 5 significant figures for 4.5039.  

6.85  ÷  112. 04 = 0.061138879 or 0.0611

The calculation above was rounded off to 3 significant figures following the number of significant figures of 6.85.

Try this!

Solve the problem below and round off to proper number of significant figures.

1.  11,254.1 g + 0.1983 g

2.  66.59 L -  3.113 L

3.  8.16 m x  5.1355 

4.  0.0154 kg ÷ 88.3 mL

Solution:

 1.   11,254.1 g

     +         0.1983  g

     __________

       11, 254.2983 g should be rounded off to 11,254.3 g

2.   66.59 L

      - 3.113 L

    _________

      63.477 L and should be rounded off to 63.48 L

3.  8.16 m x 5.1355   =  41.90568 m  or 41.9 m

4.  0.0154 kg÷ 88.3 mL  =  0.00017444054 kg/mL   or  0.000174 kg/mL

You can also change your answer in number 4 to scientific notation 1.74 x 10^-4 kg/mL.

     

Significant Figures

In calculating problems related to science or when measuring something, it is important that everyone knows significant figures.  Significant figure is a number that contains certain digits and one uncertain digit. In measuring, we approximate the last digit that is why it is called uncertain digit, at the same time when calculating we round off the last digit that is why it is uncertain digit.

Guidelines in Identifying Significant Figures

In scientific work we must be always careful in writing the proper number of significant figures.  Below are the rules in identifying significant figures.

1.  Any digit that is not zero is a significant figure.  Let say for example 315 cm shows 3 significant figures, 4,367 has 4 significant figures.

2.  Zeros in between nonzero digits are significant.  Example 2002 m has 4 significant figures, 40, 505 g has 5 significant figures.  Zeros become significant because it is in between nonzero digits.

3.  Zeros to the left of the first nonzero digit are not significant.  They are only used to determine the placement of the decimal point.  For example, 0.007 g has only one significant figure and 0.00034 has two significant figures.

4. If a number is greater that 1, zeros to the right are significant if it contains decimal point.  Example, 1.0 m has two significant figures, 20.000 cm has five significant figures.  And if the number is less that 1 only zeros to the right are significant.  For example, 0.000240 g has 3 significant figures, 0.000300 m has also 3 significant figures.

5.  For a number that does not have decimal point, the trailing zeros to right may or may not be significant.  Thus if we have 300 cm, there can be 1, 2 or 3 significant figures.  Now to avoid ambiguity, we can write the number in scientific notation.  The number 300 can be written as 3 x 10², which has 1 significant figure only, it can also be 3.0 x 10², this time has 2 significant figures and 3.00 x 10², can have 3 significant figures.


Example: 
Determine the number of significant figures in the following measurements:
a.  5.02 g
b.  0.03459 cm
c.  478 kg
d.  85.0 L
e.  1.330 x 10²² atoms

Solution:
a. There are  3 significant figures because zeros in between nonzero digits are significant,
b. There are 4 significant figures, zeros to the left are not significant.
c.  There are 3 significant figures, all non zero digits are significant.
d. Also has 3 significant figures, zeros to the right are significant if there is decimal point.
e.  There are 4 significant figures, zero to the right is significant because the number contains decimal point.


For addition, subtraction, multiplication and division of significant figures and rounding off the answers to proper number of significant figures just click HERE.

Factors that Affect Chemical Equilibrium

Chemical equilibrium means that there is a balance between the rate of forward and backward reaction.  But this condition is so sensitive that whatever changes in some factors or variables there is a change in the production of products, it can either produce more or less products.  These variables or factors are concentration, pressure, volume, and temperature.

Le Chatelier's Principle

Now in chemical equilibrium there is a general rule that will guide us in predicting the direction in which a chemical reaction will move in case concentration, pressure, volume and temperature are changed, this is called the Le Chatelier's Principle.  This was formulated by a French Chemist Henri Le Chatelier.  Le Chateliers Principle states that if an external stress is applied to a system at equilibrium the system adjust in such a way that the stress is partially offset or balance as it tries to reestablish the equilibrium.  Stress means the variables or factors that are changed, the concentration, pressure, volume and temperature.

Concentration

Concentration is the amount of solute that is present in a given solvent.  The greater the amount of the solute in the solution the higher is the concentration.  Now what happens when there is a change in the concentration of the reactants and products? Let us give an example:

FeSCN⁺²(aq)        ⇄        Fe⁺³(aq)    +   SCN^-(aq)

                                                 red                       pale yellow          colorless

The above reaction shows what happens with Iron (III) thiocyanate [Fe(SCN)3] when dissolve in water.  The solution gives a red color due to the presence of hydrated FeSCN⁺² ion.  The substance when dissolve in water gives Fe⁺³ ion and SCN^- ion.  The system of chemical equilibrium occurs.

Now, what will happen if we add substance into the solution?  Let us say sodium thiocyanate (NaSCN) is added to the solution.  In this case, the stress added to the system is the  increase in the concentration of SCN^- ion from the dissociation of NaSCN.  To offset the stress the Fe⁺³ ions will react with SCN^- ions, and the equilibrium will shift from right to left.

FeSCN⁺²(aq)     ←     Fe⁺³(aq)    +   SCN^-(aq)

The above reaction will deepen the red color.  What if iron (III) nitrate [Fe(NO₃)₃] is added?  The red color will also deepen because the concentration of Fe⁺³ ions will increase from the dissociation of Fe(NO₃)₃, and it will react with SCN^- ions, and therefore the equilibrium will shift from right to left.  Both Na⁺ ions and NO₃^- ions are colorless spectator ions.

Suppose we add oxalic acid (H₂C₂O₄) to the original solution.   Oxalic acid ionizes in water to form oxalate ion (C₂O₄⁺²), which binds strongly with Fe⁺³ ions forming a yellow ion Fe(C₂O₄)₃^-3 which removes free Fe⁺³ ions in solution. And therefore, more FeSCN⁺² units dissociate and the equilibrium shifts from left to right.  The solution will now turn to yellow because of the formation of Fe(C₂O₄)₃^-3.

FeSCN⁺²(aq)     →    Fe⁺³(aq)    +   SCN^-(aq)

Changes in Pressure and Volume

Not all solutions are affected of the change in pressure and volume.  Only gases  are the one affected because of the ability of gases to be compressed.  While liquids and gases are virtually incompressible.   Let us recall the formula for ideal gas equation:

PV = nRT

P = (n/V)RT  

The equation tells that P is inversely related to V, which means as the volume is decreased the pressure increases and vice versa.  On the other hand, the concentration, n, of the gas is directly proportional to the pressure of gas, which means if the concentration is increased the pressure also increases.

Let us have an example, suppose we have the reaction below:

N₂O₄(g)    ⇄    2NO₂(g)

The equilibrium system is in a cylinder fitted with a movable piston.  What will happen if the pressure on the gases is increased?  Increasing the pressure will result in decrease in volume and so concentration of N₂O₄ and NO₂ will increase.  Therefore, increasing pressure will cause the shift of equilibrium to the left, more N₂O₄ will form because of the decrease in volume.  Why N₂O₄?  Simply because N₂O₄ has less number moles.  In other words if pressure is decreased the equilibrium will shift to the right in favor of more number of moles NO₂.

Therefore, an increase in pressure (decrease in volume) favors the net reaction which has less number of moles, in the above reaction, it favors the backward reaction; and the decrease in pressure (increase in volume) favors the the net reaction which has more number of moles.  In the above reaction, it favors the forward reaction.

Let us have another examples:
Predict the direction of the net reaction of the following equilibrium system if pressure will be increased.
a.   2PbS(s)  +   3O₂     ⇄      2PbO(s)   +   2SO₂(g)
b.  PCl₅(g)     ⇄      PCl₃(g)   +   Cl₂(g)

Solution:  
We have to be reminded that only gases are affected by pressure because solids and liquids are incompressible.
a. There are 3 moles of reactants and 2 moles of the products not including the solids, therefore increasing pressure or decreasing the volume will cause the system to shift to the right  in favor of less than number of moles or to the product side.
b.  Here only one mole in the reactant side and two moles in the product side increasing the pressure will shift the equilibrium to the left to the reactant side with less number of moles.

Changes in Temperature

A change on concentration, pressure and volume does not alter the equilibrium constant it only alters the relative amounts of reactants and products, only change in temperature can alter the equilibrium constant.  There are two reactions possible the exothermic reaction and the endothermic reaction.  Exothermic reaction is a chemical reaction that releases heat or energy while endothermic reaction is the chemical reaction that absorbs heat.

Let us have an example,

N₂O₄(g)      ⇄     2NO₂(g)

The above reaction is endothermic in the forward reaction (absorbs heat, △H^o > 0),

heat  +  N₂O₄(g)    →   2NO₂(g)          △H^o = 58.0 kJ/mole

so the reverse reaction is exothermic (releases heat, △H^o < 0)

2NO₂(g)   →    N₂O₄(g)  +  heat      △H^o = -58.0 kJ/mole

At equilibrium, the heat effect is zero because there is not net reaction.  Increasing the temperature will favor the endothermic reaction, the forward reaction (the shift is from left to right), the amount of N₂O₄ decreases and an increase in the amount of NO₂. On the other hand a decrease in the temperature will favor the exothermic reaction reaction, the backward reaction (the shift is from right to left), which decreases the NO₂ and increases N₂O₄.

Therefore we can say that an increase in temperature will favor the endothermic reaction and the decrease in temperature favors the exothermic reaction.

The Effect of Catalyst

Catalyst enhances the chemical reaction. Adding catalyst to the reaction will not alter the chemical equilibrium constant or will not shift the equilibrium to any direction, it just increases the reaction rate by lowering the activation energy of the reacting molecules.

Adding a catalyst to a reaction that is not at equilibrium will only help the reaction to achieve equilibrium at the soonest possible time.  The same equilibrium can be obtained without a catalyst only that it takes longer time to occur.