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The percent by weight formula is a concept of Chemistry that is used to express the concentration of solutions. It is known that a composition of a given solution can be described by expressing the concentration of the given solution. The percent by weight formula is abbreviated as w/w%. It is also known as the mass percent composition. This percent by weight formula helps us calculate the ratio between solute to the solution that is, the amount of solute in a solution. In the percent by weight formula, the percent ratio is calculated for the concentration of solute to the concentration of the solution(solute + solvent). A solution is a mixture of the solid component(solute) uniformly distributed in the majority of the solvent part. The solute is present in a minor amount as compared to solvents in a solution. Let’s learn more about the topic and discuss some important questions.
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Key Takeaways: Percent by weight concept, Percent by weight formula, Concentration of solute, Solutions, Percent by weight, Molarity, Molality, Normality
Percent by Weight Formula
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There are a lot of representations for solutions, mainly molarity, molality, normality, and percent by weight formula. The most preferable and convenient method to represent solutions is by finding out the relative percent concentration of the solute dissolved in the solution. The percent by weight formula aids in the calculation of the relative percent concentration of the solute dissolved in the solution.
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How to Calculate Percent by Weight Formula?
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In order to determine the weight percent of a solution, divide the mass of solute by mass of the solution (solute and solvent together) and multiply by 100 to obtain percent.
The percent by weight formula can be expressed as:
Percent by Weight Formula Representations
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The percent by weight formula is a very reliable formula for calculating the concentration of the solute present in a solution. There are few representations of percent by weight formula-
- Percent weight by volume
The percent weight by volume gives us the weight per volume ratio. It is also abbreviated as %w/v.
- Percent weight by weight
This formula gives us the percent ratio of the weight of the molecules of the solution. This is abbreviated as %w/w.
- Percent volume by volume
The percent volume by volume gives us the percent ratio of the volume of the components present in the solution. The solute and the solvent, in this case, are in liquid form since the volume is to be considered.
Things to Remember
- The percent by weight formula is used for the calculation of the ratio of the concentration of the solute dissolved in a given solution in terms of percentage.
- The percent by weight formula can be written as %w/w. and is also known as the mass percent composition.
- The percent by weight formula is given by: Percent by weight = [grams of solute / (grams of solute + grams of solvent)] x 100 %
- The percent weight by volume is used to calculate the weight per volume ratio and is expressed as %w/v.
- A solution is a uniform liquid mixture of the solute in a given solvent where the solute is present in a lesser amount as compared to the solvent.
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Sample Questions
Ques. Calculate the grams of NaOCl (6.15% by mass) in 285 grams of a billboard bleach solution. [3 marks]
Ans. In this problem, we are required to find out the gram of the solute present in solution, so we use the percent by weight formula.
Given, mass percent = 6.15%
Mass of solution of billboard bleach solution = 285 grams.
Percentage of mass = ( mass of solute / mass of solution) x 100%
the Thus, grams of solute present in solution will be given by,
Grams of solute = (percentage of mass X gram of solution)/100
Grams of solute = (6.15 X 285) /100 = 17.52 grams.
Therefore, the gram of solute present in the billboard bleach solution is 17.52 grams.
Ques. What is the percent mass of 5g of caustic soda dissolved in 100g of water? [3 marks]
Ans. This problem can be solved with the help of percent by weight formula.
Given, mass of caustic soda = 5 grams.
Mass of water in which caustic soda is dissolved = 100 grams.
The total mass of the compound is the amount of caustic soda added to the quantity of water = (100 + 5) grams = 105 grams.
Therefore, the total mass of the compound = 105 grams.
Now, to calculate the mass percent we have the formula,
Mass percent = ( mass of solute(caustic soda) / total mass of solution) x 100%
= (5 g / 105 g) x 100%
= (5 g / 105 g) x 100%
= 0.04761 x 100%
= 4.761%.
Therefore, the mass percent of 5g of caustic soda dissolved in 100g of water is 4.761%.
Ques. Determine the percent by weight of sodium chloride in the following solution: 1.75 moles of NaCl (58.44 g/mol) dissolved in 0.550 kilograms of deionized H2O. [3 marks]
Ans. In this problem, we have to convert the units to a single comparable unit, suppose in grams.
Given,
Number of moles of NaCl = 1.75 moles.
Number of grams of deionized water = 0.550 kilograms = 550 grams.
Now, it is given that 1 mole of NaCl = 58.44 grams.
Therefore, 1.75 moles of NaCl = (58.44 x 1.75) / 1 = 102.27 grams of NaCl.
Now we substitute all the values in the percent by weight formula,
percent by weight = ( mass of solute / (mass of solute + mass of solvent) ) x 100%
= 102.27 grams / (102.27 + 550 ) grams x 100%
= 102.27 grams / 652.27 grams x 100%
= 0.1568 x 100%
= 15.68%
Ques. Calculate how many grams of NaOH and how many grams of solvent are required to make a 30.0% solution by using de-ionized water as the solvent. [3 marks]
Ans. Given,
Percent by weight of NaOH solution = 30.0%
Let x grams of solute be dissolved in 100 grams of solution.
30.0% = x g of solute in 100 grams of solution.
30.0 g of NaOH = x
Thus, 100 grams of the solution - amount of the solute = amount of solvent.
100 - 30 grams = 70 grams of solvent.
Therefore, 30 grams of NaOH are required to dissolve in 70 grams of de-ionized water as a solvent to make a 30% solution.
Ques. Determine the % of a solution that has 25.0g of NaCl dissolved in 100mL of de-ionized water. [3 marks]
Ans. We have to use the percent weight by volume to solve this problem.
Given,
Mass of solute that is NaCl = 25g
Mass/volume of solvent that is de-ionized water = 100mL equivalent to 100g.
Therefore,
Mass of solution = mass of solute + mass of solvent = 100g + 25g = 125g
Mass percent of NaCl in this solution = (25 g / 125 g) x 100% = 20%
Therefore, mass percent of NaCl in the solution is 20%.
Ques. There was 40g of sugar dissolved in 600 mL of sugar solution. Calculate the mass by volume percentage. [2 marks]
Ans. Given, mass of solute, sugar = 40 grams
Volume of solution = 600 mL equivalent to 600 g.
Therefore, according to the mass by volume formula
(Mass of solute / volume of solution ) x 100% = (40/600) x 100 = 6.66%.
Ques. A solution of ethanol in water is 10 % by volume. If the solution and pure ethanol have densities of 0.9866 g/cc and 0.785 g/cc respectively. Calculate the percent by weight approximate value. [3 marks]
Ans. We know that the volume percent is used to express the concentration of a solution when the volume of a solute and the volume of a solution is given.
And we write the equation as:
Volumepercentage=Volumeofsolute/Volumeofsolution × 100
Given,
10% of the total solution has ethanol.
Density of solution = 0.9866 g/cc
Density of pure ethanol = 0.785 g/cc
Let the total volume of solution be 100ml
Then the volume of ethanol will be 10ml.
Volume of ethanol = 10ml
Volume of solution = 100ml
Now we know, density = mass / volume
Weight of ethanol = Volume x density
= 10 x 0.785=7.85 grams
Weight of solution = 100 x 0.9866=98.66g
Weight percent of ethanol = (weight of ethanol/weight of solution) × 100%
Weight percent of ethanol = (7.85/98.66) x 100
Weight percent of ethanol = 7.95%
Ques. (a) Define the following terms : (i) Ideal solution (ii) Osmotic pressure
(b) Calculate the boiling point elevation for a solution prepared by adding 10 g CaCl2 to 200 g of water, assuming that CaCl2 is completely dissociated.
Given (Kb) for water = 0.512 K kg mol-1 ; Molar mass of CaCl2 = 111 g mol-1 (Comptt. All India 2017) [5 marks]
Ans. (a) i) An ideal solution is a solution that obeys Raoult's law for a large range of concentrations.
ii) Osmotic pressure is the minimum amount of pressure that is required to prevent the flow of pure solvent through a semipermeable membrane.
(b) Given, weight of CaCl2 = 10 grams. = w2
weight of water = 200 grams. = w1
i=3
Kb = 0.512 K kg mol-1
Molar mass of CaCl2 , m2= 111 g mol-1
The formula for calculation of boiling point elevation is
ΔTb = iKbm = iKb x [(w2 x 1000) / (m2 x w1)]
= (1000 x 0.512 x 10 x 3) / (200 x 111) = 0.69 K
ΔTb = 0.69 K
Ques. A 1.00 molal aqueous solution of trichloroacetic acid (CCl3COOH) is heated to its boiling point. The solution has the boiling point of 100.18°C. Determine the van’t Hoff factor for trichloroacetic acid. (Kb for water = 0.512 K kg mol-1) (Delhi 2012) [3 marks]
Ans. Given, molality, m = 1
T2 = 100.18°C ; T1 = 100.0°C
Therefore, ΔTb = 100.18 – 100.0 °C
i = vant Hoff factor for trichloroacetic acid.
As ΔTb= iKbm
(100.18 – 100) °C = i × 0.512 K kg mol-1 × 1 m
0.18 K = i × 0.512 K kg mol-1 × 1 m
∴ i = 0.3
Ques. Calculate the percent composition in terms of the mass of a solution obtained by mixing 300g of a 25% & 400 g of a 40% solution by mass? [3 marks]
Ans. From the given data we calculate,
mass of solute present in 400g of 40% solution = (40 / 100) x 400 grams = 160 grams.
Therefore, the total mass of solute in a mixture of the two solutions = 160+75 = 235g
Total mass of the solution mixture = 400+300 = 700g
Using the percent mass formula,
Mass percent formula = [mass of solute / total mass of the solution(solute+solvent)] x 100%
= (235 g / 700 g) x 100% = 33.57%
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