1.2.8 Virtual Lab - Mole Ratios PDF

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1.2.8 Lab 2: Mole Ratios AP Chemistry Sem 1 Points Possible: 50

Virtual Lab Assignment Name: ________________________ Date: _____________

For students unable to complete traditional bench labs during the 2020-2021 school year, the College Board is allowing virtual labs and simulations to be completed as a means to satisfy the hands-on laboratory requirement. Please complete and submit this lab assignment to your teacher as you would any other teacher-scored assignment. Adapted from: ● Quality Science Labs, Lab 2: Mole Ratios ● Christopher Becke, Introduction to Molarity and Dilutions, https://phet.colorado.edu/en/contributions/view/5385.

BACKGROUND: From the time of John Dalton’s Atomic Theory onward, chemists have understood that atoms combine in simple whole-number ratios to form compounds. The ratios are based on counting relationships between the atoms that make the compound. For example, we know ammonia has the formula, NH3, because we have, through experiments such as empirical formula determinations, counted three H atoms for each N atom in every sample of ammonia tested. Since the mole is simply a collection of like items (usually atoms, molecules, or ions), we can see that the mole ratio will be the same as the ratio by atom in the sample. In our ammonia example, the mole ratio will be 3:1—3 moles H to 1 mole N, just as ammonia’s formula gave a 3:1 H to N ratio by atom in a single molecule. Counting relationships are also evident in balanced chemical reaction equations, in which we see whole-number ratios between the participating species. Balancing a chemical reaction equation is essentially an application of the Law of Conservation of Matter (and electric charge) to obtain coefficients that express the counting relationships between species. From the coefficients, we can obtain the mole ratios. For example, in the chemical reactions in Table 2.1, the reactants combine in simple ratios. In the lab, we may encounter reactions that have different mole ratios, even when the reactants are the same. For example, combustion of CH4 in oxygen goes by different mole ratios, depending on the abundance of oxygen present: (Equation 2.1) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) (O2 abundant; 1:2 mole ratio for CH4 and O2) (Equation 2.2) 2 CH4(g) + 3 O2(g) → 2 CO(g) + 4 H2O(g) 1

(O2 not abundant; 2:3 mole ratio for CH4 and O2) Table 2.1: Chemical Reactions Ratio of Reactants

Reactants’ Ratio

(by formula unit)

(by mole)

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

1:1

1:1

Ca(NO3)2(aq) + 2 NaF(aq) → CaF2(s) + 2 NaNO3(aq)

1:2

1:2

Al(NO3)3(aq) + 3 NaOH(aq) → Al(OH)3(s) + 3 NaNO3(aq)

1:3

1:3

Reaction

Through lab measurements, we may be able to determine the mole ratios experimentally, which in turn can help us to know which chemical reaction equation applies when we attempt the chemical reaction. DIRECTIONS: In this virtual lab activity, you will engage in a guided inquiry using a simulation to learn about the concept of molarity, how to calculate molarity, and how to recognize saturated solutions. Either go directly to https://phet.colorado.edu/en/simulation/molarity , or go to https://phet.colorado.edu/, search “molarity” and select Molarity (HTML5). Have you ever made lemonade or Kool Aid with powdered drink mix? Sometimes you make it too weak and sometimes it may turn out too strong! Look at the simulation. The first slider is the “Solute Amount”. This is the amount of the solid (powder) you’re adding. The second slider is the “Solution Volume”. This is how much you’ll wind up with in your glass or pitcher. The slider to the right is “Solution Concentration”. This is a good indication of if the Kool Aid is weak, strong or just right. 1. When you first open the simulation, it gives you what looks like a pretty weak drink. What are two ways of making the drink stronger? Move just one slider to make a change that works, then hit the reset button and make a second change that works. Write your answers below: (2 points) Answer: 1) Drink can be make stronger by decreasing the volume of solvent (solution) 2) Drink can be make stronger by increasing the concentration of solute. 2

In Chemistry, we often want to do numerical calculations, so first hit “reset”, then hit the “Solution values” checkbox. You should see that the Solute Amount (moles) = 0.500 mol, the Solution Volume (Liters) = 0 .500 L, and the Solution Concentration (Molarity) = 1.000 M. 2. Play with the sliders and find other values for moles and Liters that will result in a Molarity of 1.000 M (4 points) Solute Amount (moles)

Solution Volume (Liters)

Solution Concentration (Molarity)

0.603

0.603

1.000 M

0.355

0.355

1.000 M

0.273

0.273

1.000 M

0.788

0.788

1.000 M

3. What do you notice about the values? (2 points) Answer: Same values of moles of solute and liters of solution result in 1.000M molarity of solution The concentration or strength of a solution is called its Molarity which is a ratio of the solute amount in moles divided by the solution volume in liters. We can write the formula:

Solution Concentration (Molarity)=

The units for molarity are moles per liter or

Solute Amount( moles) n ∨M = V Solution Volume(Liters)

mol L

This equation can be rearranged to solve for volume or for number of moles:

M=

n n n=M ×V V = M V

4. Using the equations, determine the Solute Amount (moles) and the Solution Volume (Liters) that would create a 0.800 molar solution of drink mix in each situation below. (4 points) moles

Liters

Molarity

Example:

0.625 L

0.800 M

0.500 mol

Show Calculations

V=

N 0.500 = =0.625 L M 0.800 3

V=

N 0.800 = =1.0 L M 0.800

0.800 mol

1.0 L

0.800 M

0.2048 mol

0.256 L

0.800 M

N=V × M =0.256 ×0.800=0.2048 mol

0.661 L

0.800 M

N=V × M =0.661× 0.800=0.5288 mol

0.831 L

0.800 M

0.5288 mole

0.665 mol

V=

N 0.665 = =0.831 L M 0.800

5. Looking at your data, for a given molarity, if you have more volume of solution, what happens to the number of moles of solute contained in that solution? (2 points) Answer: Number of moles of solute remains same but molarity is decreased by increasing the volume. Molarity Practice Calculations 6. Calculate the missing values in the table: (6 points)

M=

moles 0.887 mol

0.299 mol

Liters 0.558 L

0.830 L

Molarity 1.590 M

0.360 M

n n n=M ×V V = M V Show Calculations

V=

n 0.887 = =0.558 L M 1.590

n=M ×V =0.360 ×0.830=0.299 mol

0.750 mol

0.295 L

2.54 M

M=

n 0.750 = =2.54 M V 0.295

0.908 mol

0.585 L

1.552 M

V=

n = 0.908 =0.585 L M 1.552

4

0.205 mol

0.880L

0.233 M

0.205 mol

0.2 L

1.025 M

M=

n 0.205 = =0.233 M V 0.880

V=

n = 0.205 =0.2 L M 1.025

7. Answer True or False: (3 points) T

F

Having more moles of solute definitely means a higher concentration (molarity) solution Explain: true, because more solute moles means increase in concentration of solute.

T

F

Having a greater volume definitely means a lower concentration (molarity) solution Explain: true because more volume density becomes lower and that is why solution become less concentrated

T

F

With a given number of moles of solvent, the solution will always have the same concentration

Explain: False, if we change volume of solution with same number of moles, the concentration will be changed.

Calculating Grams If you wanted to make 0.800 Liters of a 0.531 M solution of Nickel(II) chloride, NiCl₂, how many grams of NiCl₂ would you need? Moles of nickel chloride = 0.800 × 0.531 = 0.4248 moles Mass of nickel chloride = 0.4248 × 127.87 = 54.32g A problem like this first requires you to identify how many moles of NiCl₂ using a molarity formula, and then to convert that number of moles to grams using the molar mass of NiCl₂.

5

8. First, identify your known variables (leave one blank): (1 point)

M =¿

0.45 M

n=¿

0.255mol

V =¿ 0.5 L Now, circle the version of the molarity equation that would be most easy to work with (with your unknown on the left): n= M × V

M=

n n n=M ×V V = M V

9. Next, use that equation to solve for your unknown value. (show work) (1 point) N= 0.45 × 0.5 = 0.255 moles Mass = 32.61g

Now that you know the number of moles, you will need the molar mass of NiCl₂ to calculate the grams of NiCl₂. 10. Calculate the molar mass of NiCl₂ here, showing your work: (1 point)

Molar mass = 58.69 + 35 + 35 = 128g/mole

11. Now, use the unit cancelling method to determine the grams of NiCl₂ required: (2 points) 128 0.255

mol NiC l ×

g NiC l2 32.62 mol NiC l2

In the simulation, switch the Solute from Drink Mix to Nickel(II) chloride by clicking on the box below the beaker. Adjust the sliders to check your answer with the simulation. 0.5 moles 6

0.5L volume 1.00 molarity Dilution Reset the simulation and set the quantities at 0.250 mol of Drink Mix in 0.200 L of solution. Enter the molarity in the table below. 12. Now we are going to dilute the solution only by sliding the second slider and changing the volume (adding more water). Complete the first four columns of the table, indicating if the number of moles increased, decreased or remained the same.: (11 points) moles

Liters

Molarity

Did the moles increase, decrease or stay the same?

0.250 mol

0.200 L

1.25 M

same

0.25

0.250

0.355 L

0.70 M

decreased

0.248

0.250

0.426 L

0.58 M

increased

0.247

0.250

0.520 L

0.481 M

decreased

0.250

0.250

0.726 L

0.344 M

decreased

0.249

0.250

0.881 L

0.28 M

decreased

0.246

13. Now multiply the liters (column 2) by the molarity (column 3) and enter that value in the last column. What do you notice? ( 1 point)

Number of moles decreased by increasing volume. The Molarity formula can be rearranged to solve for number of moles:

n=M ×V .

Notice that diluting the solution results in the same number of moles. Therefore, since same, the quantity M × V will remain constant.

n remains the

7

Since this is the case, we can use the following formula for dilutions:

M 1 ×V 1 =M 2 ×V 2 The initial molarity times the initial volume equals the new molarity times the new volume.

M 1 ×V 1 =M 2 ×V 2 14. Calculate the missing quantities in the table (5 points) Initial Molarity ( M₁ )

Initial Volume ( V₁ )

Final Molarity ( M₂ )

Final Volume ( V₂ )

2.000 M

0.500 L

1M

1.00 L

2.500 M

0.200 L

0.500 M

1L

0.215 M

0.785 L

0.188 M

0.897 L

1.941 M

0.306 L

0.832 M

0.714 L

2.280 M

0.368 L

1.493 M

0.562 L

Now check your answers with the simulation Saturated Solutions In the simulation, switch the Solute from Drink Mix to Copper(II) sulfate by clicking on the box below the beaker. Adjust both sliders to the bottom so that you have 0.000 moles and 0.200 L. 15. Slowly raise only the Solute Amount by (about) 0.070 moles and watch the Solution Concentration change. (3 points) moles

Molarity (M)

0.000

0.000

0.071

0.142

0.140

0.27

0.209

0.418

0.280

0.56

0.349

0.7000.84

8

0.420 16. Does the molarity continue to change? Why or why not? (1 point) Yes by changing the no. of moles, the molarity kept on changing.

17. What do you notice on the bottom of the container as you continue to add additional solute to the container? (1 point) Solution starts concentrating with increase in amount of solute.

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