Specific Gravity for Chemistry Job Exams: Everything You Need to Know

What is Specific Gravity?

In simple terms, specific gravity (SG) is the ratio of the density of a substance to the density of a reference substance, usually water at 4°C. Since the density of water is 1.0 g/mL at this temperature, the specific gravity of any substance can be thought of as a comparison to water.

The formula for Specific Gravity:

Since the density of water at 4°C is 1.0 g/mL, the formula simplifies to:

So, the specific gravity has no units. It’s a dimensionless number.

Why is Specific Gravity Important in Chemistry?

Specific gravity is important because it helps you understand the relative density of substances. If the specific gravity of a substance is greater than 1, it will sink in water (because it’s denser). If the specific gravity is less than 1, it will float. This concept is essential for tasks like separating mixtures, identifying substances, and performing quality control tests.

Real-World Examples of Specific Gravity:

• Solvents: Specific gravity helps in determining the behaviour of solvents when mixed with water.
• Metals and Minerals: Identifying rocks and minerals in geology often involves measuring specific gravity to determine their composition.
• Quality Control: In industries like pharmaceuticals and cosmetics, knowing the specific gravity of liquids and solutions ensures consistent product formulations.

How to Calculate Specific Gravity

To calculate specific gravity, you simply need to know the density of the substance and the density of water. Here's a step-by-step guide:

1. Measure the density of the substance (usually given in g/mL).
2. Use the standard density of water, which is 1.0 g/mL at 4°C.
3. Divide the density of the substance by the density of water.

For example, if a substance has a density of 2.5 g/mL, the specific gravity would be:

This means the substance is 2.5 times denser than water.

Types of Problems Related to Specific Gravity in Chemistry Exams

In chemistry job exams, you may be asked to calculate or interpret the specific gravity of various substances. These problems can vary in complexity, from simple calculations to real-world applications. Here are a few types of problems you might encounter:

1. Basic Specific Gravity Calculation

This is the most straightforward type of problem. You’re given the density of a substance and asked to calculate its specific gravity.

Example Problem:
The density of a liquid is 1.2 g/mL. What is its specific gravity?

Solution:
Since the density of water is 1.0 g/mL, you can simply divide the density of the liquid by the density of water:

The specific gravity of the liquid is 1.2.

2. Converting Specific Gravity to Density

Sometimes, you may be given the specific gravity and asked to calculate the density of a substance. This is a reversal of the previous problem.

Example Problem:
The specific gravity of a substance is 3.0. What is its density?

Solution:
To find the density, multiply the specific gravity by the density of water:

So, the density of the substance is 3.0 g/mL.

3. Comparing the Density of Substances

Another common problem involves comparing the specific gravity of two substances and determining which one is denser.

Example Problem:
Substance A has a density of 1.5 g/mL, and Substance B has a density of 0.8 g/mL. Which substance is denser?

Solution:
The specific gravity of Substance A is:

The specific gravity of Substance B is:

Since Substance A has a higher specific gravity, it is denser than Substance B.

4. Density of a Mixture

You may be asked to calculate the specific gravity of a mixture based on the densities and volumes of its components.

Example Problem:
You mix 50 mL of a substance with a specific gravity of 1.2 and 50 mL of a substance with a specific gravity of 0.9. What is the specific gravity of the mixture?

Solution:
First, calculate the masses of each substance:

• Mass of Substance 1 = Volume × Density = 50 mL × 1.2 g/mL = 60 g
• Mass of Substance 2 = Volume × Density = 50 mL × 0.9 g/mL = 45 g

The total mass is 60 g + 45 g = 105 g, and the total volume is 50 mL + 50 mL = 100 mL.

Now, calculate the density of the mixture:

Finally, calculate the specific gravity:

So, the specific gravity of the mixture is 1.05.

5. Calculating Mass from Specific Gravity and Volume

In some cases, you may know the specific gravity and the volume of a substance and need to find its mass. This is commonly encountered in chemical processes where exact quantities are necessary.

Example Problem:
A liquid has a specific gravity of 0.85, and you have a 500 mL sample of it. What is the mass of this liquid?

Solution:
To find the mass, use the formula:

$$\text{Mass}=\text{Specific Gravity}\times \text{Density of Water}\times \text{Volume}$$

Since the density of water is 1.0 g/mL, we can simplify this to:

$$\text{Mass}=SG\times \text{Volume}\text{<br>}\text{<br>}\text{<br>}\text{<br>}\text{<br>}\text{<br>}$$$$\text{<br>}\text{Mass}=0.85\times 500\text{mL}=425\text{g}$$

So, the mass of the liquid is 425 g.

6. Finding Volume from Specific Gravity and Mass

Another common problem is to determine the volume of a substance if you know its specific gravity and mass. This can be especially useful when working with storage containers or when measuring out materials for a reaction.

Example Problem:
A sample has a mass of 200 g and a specific gravity of 1.25. What volume will this sample occupy?

Solution:
Rearrange the specific gravity formula to solve for volume:

$$\text{Volume}=\frac{\text{Mass}}{\text{Specific Gravity}\times \text{Density of Water}}$$

Since the density of water is 1.0 g/mL, this becomes:

$$\text{Volume}=\frac{\text{Mass}}{S\mathrm{G}}$$$$\text{Volume}=\frac{200\text{g}}{1.25}=160\text{mL}$$

So, the volume of the sample is 160 mL.

7. Temperature-Adjusted Specific Gravity Calculations

Specific gravity can vary with temperature, so you may encounter problems requiring adjustments for changes in temperature. This is because density changes with temperature, and so does specific gravity.

Example Problem:
The specific gravity of a liquid is given as 1.1 at 20°C. If the temperature is raised to 30°C, the density decreases by 2%. What is the new specific gravity?

Solution:
First, calculate the new density:

$$\text{New Density}=\text{Original Density}\times (1-\text{Percentage Decrease})$$

Since specific gravity was originally 1.1, the original density of the liquid is:

$$\text{Density of Liquid}=1.1\times 1.0\text{g/mL}=1.1\text{g/mL}$$

Now, apply the 2% decrease:

$$\text{New Density}=1.1\text{g/mL}\times (1-0.02)=1.1\times 0.98=1.078\text{g/mL}$$

Finally, calculate the new specific gravity:

$$SG=\frac{\text{New Density}}{\text{Density of Water}}=\frac{1.078\text{g/mL}}{1.0\text{g/mL}}=1.078$$

So, the new specific gravity at 30°C is 1.078.

8. Dilution Problems Involving Specific Gravity

In some applications, you may need to dilute a solution to achieve a specific concentration or specific gravity, especially in quality control and lab work.

Example Problem:
You have 200 mL of a solution with a specific gravity of 1.2. You want to dilute it with water to reach a specific gravity of 1.1. How much water should you add?

Solution:
Since specific gravity is directly proportional to concentration, this is a ratio problem. Set up the ratio:

$$\frac{\text{SG}_{\text{initial}} \times V_{\text{initial}}}{\text{SG}_{\text{final}}} = V_{\text{final}}$$ $$\frac{1.2 \times 200 \, \text{mL}}{1.1} = V_{\text{final}}$$ $$V_{\text{final}} = \frac{240}{1.1} = 218.18 \, \text{mL}$$

Now, calculate how much water to add:

$$\text{Water to Add} = V_{\text{final}} - V_{\text{initial}}$$ $$\text{Water to Add}=218.18\text{mL}-200\text{mL}=18.18\text{mL}$$

So, you should add about 18.18 mL of water to achieve the desired specific gravity.

9. Specific Gravity and Percent Concentration Calculation

Some problems might ask you to calculate per cent concentration using specific gravity, particularly in cases involving weight/volume or weight/weight concentrations.

Example Problem:
A solution has a specific gravity of 1.5, and it contains 30% (w/w) solute. Calculate the concentration in g/mL.

Solution:

1. First, calculate the mass of solute per mL using specific gravity: $$\text{Mass of Solution} = 1.5 \, \text{g/mL}$$
   
2. Now, calculate the mass of solute in 1 mL of solution: $$\text{Mass of Solute} = \text{Mass of Solution} \times \text{Percent Concentration}$$
   $$\text{Mass of Solute}=1.5\text{g/mL}\times 0.30=0.45\text{g/mL}$$

So, the concentration of solute is 0.45 g/mL.

10. Problem with Unknown Specific Gravity from Mass and Volume

Sometimes, you’ll be given the mass and volume of a substance and need to determine its specific gravity, which can be useful in identifying unknown samples.

Example Problem:
A sample weighs 300 g and occupies a volume of 250 mL. What is its specific gravity?

Solution:
Calculate the density of the sample first:

$$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$$ $$\text{Density} = \frac{300 \, \text{g}}{250 \, \text{mL}} = 1.2 \, \text{g/mL}$$

Now, calculate the specific gravity:

$$SG=\frac{\text{Density of Substance}}{\text{Density of Water}}=\frac{1.2\text{g/mL}}{1.0\text{g/mL}}=1.2$$

So, the specific gravity of the sample is 1.2.

11. Mixing Solutions with Different Specific Gravities

When mixing two solutions with different specific gravities, you might need to calculate the specific gravity of the final mixture. This is common in QC settings where solutions or reagents are blended.

Example Problem:
You mix 300 mL of Solution A (specific gravity 1.2) with 200 mL of Solution B (specific gravity 0.9). What is the specific gravity of the final mixture?

Solution:

1. First, calculate the total mass of each solution using $\text{Mass} = \text{Specific Gravity} \times \text{Volume}$
   

   • Mass of Solution A: $1.2 \times 300 = 360 \, \text{g}$
     
   • Mass of Solution B: $0.9 \times 200 = 180 \, \text{g}$
     

2. Find the total mass and total volume of the mixture:

   • Total Mass: $360 + 180 = 540 \, \text{g}$
   • Total Volume: $300 + 200 = 500 \, \text{mL}$
     

3. Now, calculate the specific gravity of the mixture:

   $$\text{Specific Gravity of Mixture}=\frac{\text{Total Mass}}{\text{Total Volume}}=\frac{540\text{g}}{500\text{mL}}=1.08$$

So, the specific gravity of the mixture is 1.08.

12. Impurity or Contamination Impact on Specific Gravity

Sometimes, you’ll be asked how an impurity affects the specific gravity of a substance. This is crucial in quality control because impurities often alter physical properties.

Example Problem:
A solution has a specific gravity of 1.15. Due to contamination, its mass increases by 5%, but its volume remains the same. What is the new specific gravity?

Solution:

1. Calculate the new mass using the 5% increase.

   • Original mass (assuming 1 mL volume for simplicity): $1.15 \times 1 = 1.15 \, \text{g}$
     
   • New mass: $1.15 \, \text{g} \times 1.05 = 1.2075 \, \text{g}$
     

2. Since the volume is unchanged, the new specific gravity is:

   $$\text{New Specific Gravity}=\frac{\text{New Mass}}{\text{Volume}}=\frac{1.2075\text{g}}{1\text{mL}}=1.2075$$

The new specific gravity is 1.2075.

13. Converting Between Specific Gravity and Percentage Concentration

This type of problem involves converting between specific gravity and weight/weight or weight/volume percentage concentrations, often for formulations.

Example Problem:
A 20% (w/w) NaCl solution has a specific gravity of 1.14. What is the mass of NaCl in 100 mL of this solution?

Solution:

1. Calculate the mass of the solution:

   $$\text{Mass of Solution}=\text{Specific Gravity}\times \text{Volume}=1.14\times 100\text{mL}=114\text{g}$$

2. Since it’s a 20% (w/w) solution, the mass of NaCl is:

   $$\text{Mass of NaCl}=0.20\times 114\text{g}=22.8\text{g}$$

So, 100 mL of the solution contains 22.8 g of NaCl.

14. Calculating Specific Gravity from Solution Density with Known Solute

When you have the density of a solution and the percentage of solute, you may need to find the specific gravity to determine if the solution meets QC standards.

Example Problem:
A sugar solution has a density of 1.12 g/mL and contains 15% (w/w) sugar. What is the specific gravity?

Solution:

1. The specific gravity is simply the density of the solution relative to water. $$SG=\frac{\text{Density of Solution}}{\text{Density of Water}}=\frac{1.12\text{g/mL}}{1.0\text{g/mL}}=1.12$$

The specific gravity of the sugar solution is 1.12.

15. Buoyancy and Floating Objects

In some cases, specific gravity helps determine if an object will float or sink, useful in material selection or density testing in QC.

Example Problem:
An object has a specific gravity of 0.95. Will it float or sink in water?

Solution: Since its specific gravity (0.95) is less than 1, it is less dense than water and will float.

16. Percentage Error in Specific Gravity Measurements

In QC, measuring errors are common. Problems may ask you to calculate the percentage error in specific gravity measurements.

Example Problem:
A QC analyst measured the specific gravity of a solution to be 1.25, but the true value is 1.30. What is the percentage error?

Solution:

1. Calculate the error:

   $$\text{Error}=\text{True Value}-\text{Measured Value}=1.30-1.25=0.05$$

2. Calculate the percentage error:

   $$\text{Percentage Error}=\frac{\text{Error}}{\text{True Value}}\times 100=\frac{0.05}{1.30}\times 100\approx 3.85\mathrm{\%}$$

So, the percentage error is approximately 3.85%.

17. Determining Purity Using Specific Gravity

This is especially relevant in the quality control of raw materials. If the specific gravity deviates from the expected value, it can indicate impurities.

Example Problem:
A sample of ethanol has a specific gravity of 0.8, but pure ethanol has a specific gravity of 0.789. Calculate the approximate percentage of impurity in the sample.

Solution:

1. Calculate the deviation:

   $$\text{Deviation}=\text{Measured SG}-\text{Pure SG}=0.8-0.789=0.011$$

2. Approximate the percentage impurity by comparing the deviation to the pure specific gravity:

   $$\text{Percentage Impurity}\approx \frac{\text{Deviation}}{\text{Pure SG}}\times 100=\frac{0.011}{0.789}\times 100\approx 1.39\mathrm{\%}$$

The sample has approximately 1.39% impurity.

18. Volume Displacement and Specific Gravity

Sometimes, you may calculate specific gravity by determining how much water an object displaces, an indirect but useful approach in quality control.

Example Problem:
A metal object weighs 500 g in air and displaces 100 mL of water. What is its specific gravity?

Solution:

1. Use the displaced water volume to find the density of the object:

   $$\text{Density of Object}=\frac{\text{Mass}}{\text{Displaced Volume}}=\frac{500\text{g}}{100\text{mL}}=5\text{g/mL}$$

2. Calculate the specific gravity:

   $$SG=\frac{\text{Density of Object}}{\text{Density of Water}}=\frac{5\text{g/mL}}{1.0\text{g/mL}}=5$$

The specific gravity of the object is 5.