LAB#2
LabExercise2-Density_100AAdone xCalculations__AADONE.xlsx
F
L
C
Chem 305 Lab Exercise #2 – Density
FLC Chem 305 Lab Exercise #2 – Density
FLC Chem 305 Lab Exercise #2 – Density
Folsom Lake College – Chemistry 305 Laboratory Exercise #2
Density
Your Name:
Date:
Partner’s Name: Lab Section: M
Tu Wed Th (circle one)
Introduction: Density (d or the Greek letter ρ – rho) is defined as mass per unit volume and is a physical property of a substance. Physical properties of a substance are those that describe the substance without causing change to its chemical composition. Color, odor, density, melting and boiling points, hardness, metallic luster, ductility, malleability and viscosity are all physical properties.
Physical properties are not affected by quantity. For example, both one drop of water or the water of an entire swimming pool boils at 100°C at sea level. And because density is a physical property, one
g
ram of gold has the same density as 100 g of gold.
g
The density of solids is typically expressed as cm3 , and the density of liquids is typically expressed as g 3 = 1 mL. The density of
, although these units are sometimes used interchangeably since 1 cm mL g gasses is typically expressed as .
L
To better grasp the concept of density consider the following: If the volume of two different substances, call one substance A and one substance B, are equal, but A is heavier than B, then the density of A is more than the density of B.
Objective: The objective of this laboratory exercise is to introduce the concept of density and to accurately plot generated data as a standard curve. The standard curve will then be used to determine the % salt in a solution of unknown concentration.
Procedure: To determine the density and concentration of a salt water solution of unknown concentration, students will first calculate the densities of four standard salt solutions by weighing a known volume of each standard. Once determined, the calculated densities will be plotted as a function of their given concentration to create a standard curve. The concentration of the unknown salt solution can be determined by extrapolation along the standard curve.
Step 1: Weigh your dry, empty graduated cylinder and record the mass.
Step 2: Fill the cylinder with more than 10 mL of distilled water. Record the volume to the nearest 0.1 mL and record the mass. Hint: The volume used is not important, but just be sure to record the volume used exactly.
Step 3: Empty the water from the graduated cylinder. Now, to ensure the next solution in the graduated
cylinder is not diluted by the previous solution, add approximately 5 mL of the 5% salt solution to the cylinder, swirl the solution around and discard. Repeat this step one more time to ensure that all the liquid adhering to the side of the graduated cylinder is now the 5% salt solution.
Step 4: Fill the cylinder with a minimum of 10 mL of the 5% salt solution. Record the mass and volume of this solution to the nearest 0.1 mL.
Step 5, 6, and 7: Repeat steps 2 – 4 above for each of the 10% and 15% salt solutions. Determine the density of each of these solutions using the formula for density given above by substituting in the mass and volume you measured for each corresponding salt solution.
Step 8: Plot your standard curve. Be sure to follow the graph plotting guidelines outlined below.
Step 9: Obtain approximately 20 mL of the unknown salt solution. Record the identification letter. Repeat Steps 2 – 4 above to determine the density of the unknown solution.
Step 10: Determine the percent concentration of the unknown salt solution by extrapolation using your standard curve. To do this, find the calculated density of your unknown salt solution on the yaxis of your graph. Next, move horizontally across the graph until you reach the diagonal line of your standard curve. From the point where your horizontal line intersects your standard curve, draw a vertical line down to the x-axis. The corresponding concentration of your unknown is the value that corresponds to the point where the vertical line intersects the x-axis.
Graph Plotting Guidelines:
1. Determine the scale of your graph by deciding how many units each line on the graph paper represents. The calculated density values (y-axis) range can be determined by taking the difference between your highest and lowest value and dividing the resulting number by the number of graduations on your graph paper. This will give you the increment value for each graduation. Then start your y-scale at your lowest value and label each major graduation by adding the appropriate increment to your lowest value. The percent salt concentration (x-axis) will range between 0 and 15%. To achieve the best precision and accuracy, spread the values far apart along the x-axis and yaxis in order to have your graph cover as much of the graph paper as possible.
2. Determine the starting values for both vertical and horizontal axes. It is not always desirable to begin every graph at zero.
3. Number and mark off the major divisions along each axis.
4. Plot the percent salt composition (0%, 5%, 10%, and 15%) along the x-axis (horizontal axis) on the long edge of your graph paper.
5. Plot your calculated densities for each of these solutions along the y-axis (vertical axis) on the short edge of your graph paper.
6. Label each axis with a description of the units (such as density (g/mL) or % salt concentration).
7. Place a point on your graph representing the calculated density of your solution vs. percent concentration at the intersection of the horizontal and vertical lines for each of your solutions.
8. Draw a straight “line of best fit” using a ruler that gets as close as possible to as many points as possible. This line now represents a “standard curve” from which you can determine the percent composition of your unknown.
9. Determine the percent composition of the unknown by interpolation. Interpolation is when one of the variables of the unknown are known (in this case, density). Locate the density of the unknown on the y-axis, then extend out in an imaginary horizontal straight line to where you intersect the “standard curve”. As soon as you intersect the standard curve, drop a vertical line straight down to the x-axis. The point at the x-axis represents the percent salt of the unknown.
(Show calculations in the spaces below) Mass of dry graduated cylinder 46.393g
Mass of cylinder + distilled H2O: 61.187g
Mass of H2O: 14.794 g
Volume of H2O:15mL
Density of H2O: 0.98626667g/mL
Mass of cylinder + 5% salt solution:63.029g
Mass of 5% salt solution:16.636g
Volume of 5% salt solution:16mL
Density of 5% salt solution: 1.03975 g/mL
Mass of cylinder + 10% salt solution: 63.036g
Mass of 10% salt solution:16.643g
Volume of 10% salt solution:15.5mL
Density of 10% salt solution: 1.073741935 g/mL
Mass of cylinder + 15% salt solution:62.847g
Mass of 15% salt solution: 16.454g
Volume of 15% salt solution:15.0mL
Density of 15% salt solution:1.096933333 g/mL
Mass of cylinder + unknown salt solution:64.365g
Mass of unknown salt solution:17.972g
Volume of unknown salt solution:17.0mL
Density of unknown salt solution: 1.057176g/mL
Identifying letter of unknown salt solution L
Percent salt in your unknown solution (from graph) :8.2%
Density Problems:
1.
A. Calculate the density of an irregular shaped metal object that has a mass of 321 g and a volume of
45.2 cm3.
Answer: 7.10g/cm3 (2 decimal place)
The formula for density is; density= mass/ volume
So, the density of an irregular shaped metal object can be computed as follows;
Density=321g/45.2 cm3
= 7.10176991 g/cm3
B. Based on the information provided in the table below, what is the identity of the metal?
Metal
Density (g/mL)
Metal
Density (g/mL)
Magnesium
1.7
Nickel
8.9
Aluminum
2.7
Copper
9.0
Zinc
7.1
Silver
10.5
Tin
7.3
Lead
11.4
Iron
7.9
Gold
19.3
Answer: The metal that corresponds to the density computed in (A) above is Zinc
2. What is the volume of 4.00 g of air if the density of air is 1.19 g/L?
Answer: 3.36134454 L
Workings
The formula for volume is: Volume =mass/density
Given, mass = 4.00g , density= 1.19 g/L
So, the volume of the air;
Volume of air =4.00g/1.19 g/L
=3.36134454 L
3. Calculate the mass of a quart (946 mL) of mercury. The density of mercury is 13.6 g/mL.
Answer: 12,865.6 g
workings
The formula for mass is; mass=density* volume
Given volume=946mL,density=13.6 g/mL.
So, the mass of a quart (946 mL) of mercury =(13.6 g/mL) *(946 mL)= 12,865.6 g
4. A student performed the experiment but varied the volumes used when weighing out the liquids. For the 0% salt solution, the student used 13.2 mL, for the 5% solution the student used 12.9 mL and so on. Assuming everything else was performed precisely and accurately, will the differing volumes affect the outcome of the experiment? Explain your answer.
Answer: Differing the volumes doesn’t affect the outcome of the experiment. The volume used is not important as long as it was recorded properly, regardless of the volume the density should remain the same.
5.
What did you learn from this experiment?
Answer: From this experiment, I learned the densities of salt solutions is primarily dependent on the concentration and the volume of the salt solution. Notably, the density will increase with the increase of concentration.
Standard curve
2.1
2.1
2.1
Sheet1
| Mass | 46.393 | ||||
| 0% salt soln | 5% salt soln | 10% salt soln | 15% salt soln | Unknown salt soln | |
| Volume(mL) | 15.0 | 16.0 | 15.5 | 17.0 | |
| Mass (g) | 14.794 | 16.636 | 16.643 | 16.454 | 17.972 |
| Density | 0.9862666667 | 1.03975 | 1.0737419355 | 1.0969333333 | 1.0571764706 |