molar mass. do now you have a bag of blue and red marbles. you’re trying to figure out the mass of...

Molar Mass

Do Now

• You have a bag of blue and red marbles. You’re trying to figure out the mass of the bag, but you don’t have a balance/ scale.

• What might you need to know to solve this problem? How might you be able to find the mass?

Do Now- con’t.

• Your bag contains 10 blue marbles and 15 red marbles.

• What else may we need to know?

Do Now- con’t.

• Your bag contains 10 blue marbles and 15 red marbles.

• Each blue marble weighs 2 grams, and each red marbles weighs 3 grams.

• What is the mass of the bag?

Mass on the Molecular Level

• Just like we weren’t able to directly weigh the bag of marbles, it’s impossible to “weigh” one molecule.– But, we can find the mass of a molecule by adding up

the mass of its individual atoms

• Molar Mass: the mass (in grams) of one mole of molecules – This is based on the atomic mass in the periodic

table

Determining Molar Mass1. From the formula, write down each of the different

elements that are present.MgCl2: Mg

Cl

2. Identify how many atoms of each element are present using the subscripts (no subscript = just one atom)

Mg: 1Cl: 2

3. Multiply the number of atoms of each element by that element’s atomic mass, then find the total mass by adding all the elements’ masses together

Mg: 1 x 24.3050 g = 24.3050gCl: 2 x 35.4527 g = + 70.9054g

95.2104g

Practice!

• Determine the molar mass of H2SO4 (sulfuric acid).

Practice!

• Determine the molar mass of H2SO4 (sulfuric acid).

H2SO4

H: 2 x 1.0079g = 2.0158gS: 1 x 32.066g = 32.066gO: 4 x 15.999 = 63.996g

98.0778g

Practice!

• What is the molar mass of (H3O)3PO4?

Practice!


• Note: when you have parentheses in the formula, the subscript outside them is distributed (multiplied) by all of the subscripts inside!

Practice!


H: (3 x 3) = 9O: (1 x 3) + 4 = 7P: 1

H: 9 x 1.0079 = 9.0711O: 7 x 15.999 = 111.993 P: 1 x 30.974 = 30.974

152.038g

What exactly is a mole?

• Avogadro’s Number: 6.022 x 1023 • The number of carbon atoms in exactly 12g of pure

12C

• 1 mole of ANYTHING = 6.022 x 1023 units of that substance

• How many teachers are in 1 mole of teachers?• How many pencils are in 1 mole of pencils?• How many candy bars are in 1 mole of candy bars?

What exactly is a mole?• Avogadro’s Number: 6.022 x 1023 • The number of carbon atoms in exactly 12g of pure 12C

• 1 mole of ANYTHING = 6.022 x 1023 units of that substance

• How many teachers are in 1 mole of teachers?• How many pencils are in 1 mole of pencils?• How many candy bars are in 1 mole of candy bars?

6.022 x 1023!!!

If we know moles…• The molar mass is the grams of that substance that it

takes to make up 1 mole• So, if we know how many moles we started with, we

can determine the mass of our substance

• OR,

• We also know the number of particles in one mole (Avogadro’s number)

• So, we can determine the number of molecules of our substance

Our map…

Grams Moles # of Particles

Molar MassAvogadro’sNumber

Going between Grams and Moles

We have 24g of carbon. We know that the molar mass is 12.011g/mol (there are 12.011 grams of carbon in one mole)

• SET UP THE CONVERSION! Start with what you know. The molar mass is our conversion factor.


We have 24g of carbon. We know that the molar mass is 12.011g/mol (there are 12.011 grams of carbon in one mole)

• SET UP THE CONVERSION! Start with what you know. The molar mass is our conversion factor.

24g C x 1 mole = 24g C mol = 1.998 mol C1 12.011g C 12.011g C


• We have 3 moles of sulfur. How many grams do we have?


• We have 3 moles of sulfur. How many grams do we have?

3 mol S x 32.066g S = 96.198 g S1 1 mol S

Higher Order Questions

• What is another real-life application of stoichiometry?

• What is happening on a molecular level when two substances react to form new products?

THINK-WRITE-PAIR-SHARE!Pick one topic to contemplate with your partner.

CO2(g) + 2LiOH(s) -> Li2CO3(s) + H2O(l)

• In a spacecraft, the carbon dioxide exhaled by astronauts can be removed by its reaction with lithium hydroxide, LiOH, according to the above chemical equation.

How many grams of carbon dioxide are produced if the average person breathes out 20 mol of CO2 per day?

You Practice!

Convert from grams to moles:• 125g phosphorus (P)• 75g NaCl

Convert from moles to grams:• 5 moles of carbon (C)• 7 moles of HBr

You Practice!

Convert from grams to moles:• 125g phosphorus (P) = 4.04 mol P• 75g NaCl = 1.28 mol NaCl

Convert from moles to grams:• 5 moles of carbon (C) = 60.055g C• 7 moles of HBr = 566.4g HBr

Going Between Moles and Particles

• We have 15 x 1023 carbon atoms. How many moles is this?

• SET UP THE CONVERSION! Start with what you know. Avogadro’s number is our conversion factor.


• We have 15 x 1023 carbon atoms. How many moles is this?

• SET UP THE CONVERSION! Start with what you know. Avogadro’s number is our conversion factor.

15 x 1023 C atoms x 1 mol

6.022 x 1023 atoms

= 2.49 mol C


• We have 5.5 mol of NaCl. How many molecules is this?


• We have 5.5 mol of NaCl. How many molecules is this?

5.5 mol NaCl x 6.022 x 1023 molecules

1 mol= 33.121 x 1023 molecules NaCl= 3.3121 x 1024 molecules NaCl (correct scientific

notation)

Going Between Grams and Particles

• This is just another conversion! We don’t know the relationship between grams and particles, so what could we go to first?


• This is just another conversion! We don’t know the relationship between grams and particles, so what could we go to first?

• MOLES!!! Moles are like our base in metric conversions- when in doubt, convert to moles!


• We have 600 x 1023 carbon atoms. How many grams is this?


• We have 600 x 1023 carbon atoms. How many grams is this?

600 x 1023 atoms C x 1 mol C x 6.022 x 1023 atoms C

12.011g C= 119.7g C1 mol C


• We have 30g of pure copper. How many atoms of copper is this?


• We have 30g of pure copper. How many atoms of copper is this?

30g Cu x 1 mol Cu x 6.022 x 1023 atoms Cu = 63.546g Cu 1 mol Cu

= 2.8 x 1023 atoms Cu

Density

• Density: a physical property of a substance, representing the mass per unit volume

d = m/Vm = mass (in grams)V = volume (in mL)

Note: 1mL = 1cm3

We can use the density to determine the mass if it isn’t given!

Using Density to Determine Mass

• Determine the mass of mercury (Hg) in 30mL if the density of mercury is 13.55g/mL.Set up the equation! Fill in what you know, then

solve.


• Determine the mass of mercury (Hg) in 30mL if the density of mercury is 13.55g/mL.

d = m/V13.55g = m

mL 30mL

13.55g x 30 = m

406.5g = m


• Turpentine has a density of 0.85g/mL. How many grams are in 75mL?


• Turpentine has a density of 0.85g/mL. How many grams are in 75mL?

d = m/V0.85g = m

mL 75mL

0.85g x 75 = m

63.75g = m

Exit Ticket #1

Exit Ticket #1Solve the following problems, showing all work for full

credit.1. What is the molar mass of acetic acid (HC2H3O2)? (1

point)2. How many pennies are in one mole of pennies? (1

point) 3. How many moles are in 50g of magnesium? Show

your work! (2 points)4. How many grams are in 3.5mol of CaO? Show your

work! (3 points)5. How many molecules are in 15g of HF? Show your

work! (3 points)

Percent Composition

• You always bring a lunch to school. You wanted to know how heavy your lunch was, so you decided to weigh each of the foods you packed. Your sandwhich was 550g. Your apple was 200g. Your cookie was 50g. Your yogurt was 200g. What percent of the mass of your lunch was from the sandwich?

Percent Composition• Percent Composition: a given element’s portion

of the molecule’s mass = molar mass of the element x

100%molar mass of the whole molecule

1. Find the mass of that element in the molecule2. Find the total mass of the molecule3. Enter values into the above equation

**If you add up the percentages for all elements in the compound, you should get 100%

Percent Composition: Example

• What percent of the molecule’s mass is hydrogen in H2SO4?

Percent Composition: Example• What percent of the molecule’s mass is hydrogen

in H2SO4?

H: 2 x 1.0079g = 2.0158gS: 1 x 32.066g = 32.066gO: 4 x 15.999g = 63.996g

= 98. 078g

g H x 100% = 2.0158g x 100% = 0.0206 x 100%g H2SO4 98.078g

= 2.06%

Percent Composition: Example 2

• Still using H2SO4, determine the percentage of the mass for sulfur and oxygen.

Percent Composition: Example 2

• Still using H2SO4, determine the percentage of the mass for sulfur and oxygen.

• S: 32.066g/ 98. 078g x 100% = 32.69% S

• O: 63.996g/ 98. 078g x 100% = 65.25% O

Empirical Formulas• Empirical Formula: the simplest whole-number ratio of

atoms in a compound

1. Assume that you have 100g of substance2. Determine the number of grams of each element from

the percent composition (ex. 50% carbon = 50g carbon)3. Convert each element from grams to moles4. Divide the moles of each element by the smallest value to

get the simplest whole-number ratio• If you have a decimal of 0.5, multiply EVERYTHING by 2• If you have a decimal of 0.333, multiply EVERYTHING by 3• If you have a decimal of 0.25, multiply EVERYTHING by 4• If the number is VERY close to a whole number (ex. 0.999), it is

acceptable to round

Empirical Formula Example

• Determine the empirical formula for a compound consisting of the following:

71.65% Cl 24.27% C 4.07% H

Empirical Formula Example

• Determine the empirical formula for a compound consisting of the following:

71.65% Cl 24.27% C 4.07% HCl: 71.65g x 1mol/35.45g = 2.021mol/ 2.021 = 1C: 24.27g x 1mol/ 12.01g = 2.021mol/ 2.021 = 1H: 4.07g x 1mol/1.008g = 4.04mol/ 2.021 = 2

ClCH2

Molecular Formula• Molecular Formula: the exact formula of a

molecule– May or may not be the same as the empirical formula– You need to know the molar mass of the compound!

1. Determine the empirical formula2. Find the molar mass of the empirical formula3. Divide the molar mass of the compound by the

molar mass from the empirical formula4. Multiply the subscripts in the empirical formula

by the above ratio

Molecular Formula Example

• Earlier, we determined the empirical formula was ClCH2. Given that the molar mass of the compound is 98.96g/mol, determine the molecular formula.

Molecular Formula Example

• Earlier, we determined the empirical formula was ClCH2. Given that the molar mass of the compound is 98.96g/mol, determine the molecular formula.

• ClCH2: 35.45g + 12.01g + 2.02g = 49.48g/mol• 98.96g/mol = 2

49.48g/mol• (ClCH2)2 = Cl2C2H4

Practice

• Determine the empirical and molecular formulas for a compound containing 40.0% carbon, 6.7% hydrogen, and 53.5% oxygen. The compound has a molar mass of 180.156g/mol.

PracticeDetermine the empirical and molecular formulas for a compound

containing 40.0% carbon, 6.7% hydrogen, and 53.5% oxygen. The compound has a molar mass of 180.156g/mol.

C: 40.0g x 1mol/12.0107g = 3.33mol/ 3.33 = 1H: 6.7g x 1mol/ 1.00794g = 6.65mol/ 3.33 = 2O: 53.3g x 1mol/15.9994g = 3.33mol/ 3.33 = 1

Empirical: CH2O = 30.026g/mol

180.156 = 6 : CH2O Molecular: C6H12O6

30.026

Exit Ticket #2

Exit Ticket #2- Show ALL Work!1. What is the percent composition of oxygen in H2O? (2

points)2. Which of the following is an empirical formula? (1 point)

a. C6H12O6 b. C2H4NO c. C2H6 d. All of the above.

3. A compound’s empirical formula is C3H7 and its molecular weight is 86 g/mole. Find its molecular formula. (2 points)

4. A compound consists of 72.2% magnesium and 27.8% nitrogen by mass. What is its empirical formula? (2 points)

5. A compound is 58.8 % C, 9.9 % H, and 31.3 % O. If its molecular mass is 306 g/mole, what is the molecular formula? (3 points)

Exit Ticket #3

Exit Ticket #3

• Balance the following reactions. Write a coefficient 1 wherever needed so I can tell you didn’t just leave it blank. (2 points each)

1. S8 + NO2 SO2 + N2

2. C3H8 + O2 CO2 + H2O

3. C6H6 + HNO3 C6H5NO2 + H2O

4. C5H10 + O2 CH2O

5. N2 + C2H6 N2H4 + C2H2

Stoichiometry

• Stoichiometry: the area of chemistry that considers the quantities of materials consumed and produced in chemical reactions

• The coefficients in a balanced reaction tell you the RATIO of MOLES for each compound that are needed for the reaction

Reading the Reaction Equation

• Cl2 + 2 KBr 2 KCl + Br2

– How many moles of Cl2 will react with 2 moles of KBr?

• P4 + 5 O2 + 6 H2O 4 H3PO4 – How many moles of O2 are needed to produce 4

moles of H3PO4?

• 2 Fe + 3 Cl2 2 FeCl3

– How many moles of Fe will produce 2 moles of FeCl3?

Reading the Reaction Equation

• Cl2 + 2 KBr 2 KCl + Br2

– How many moles of Cl2 will react with 2 moles of KBr? 1 mole

• P4 + 5 O2 + 6 H2O 4 H3PO4 – How many moles of O2 are needed to produce 4

moles of H3PO4? 5 moles

• 2 Fe + 3 Cl2 2 FeCl3

– How many moles of Fe will produce 2 moles of FeCl3? 2 moles

Stoichiometry

P4 + 5 O2 + 6 H2O 4 H3PO4

What is the mole ratio between O2 and H3PO4? (How many moles of O2 per how many moles of H3PO4?)

Stoichiometry

P4 + 5 O2 + 6 H2O 4 H3PO4

What is the mole ratio between O2 and H3PO4? (How many moles of O2 per how many moles of H3PO4?)

There are 5 mol of O2 per 4 mol of H3PO4

5 mol O2 or 4 mol H3PO4

4 mol H3PO4 5 mol O2

Stoichiometry

P4 + 5 O2 + 6 H2O 4 H3PO4

What is the mole ratio between P4 and O2?

What is the mole ratio between P4 and H2O?

What is the mole ratio between P4 and H3PO4?

Stoichiometry

P4 + 5 O2 + 6 H2O 4 H3PO4

What is the mole ratio between P4 and O2?

1mol P4/ 5 mol O2

What is the mole ratio between P4 and H2O?

1mol P4/ 6mol H2O

What is the mole ratio between P4 and H3PO4?

1mol P4/4mol H3PO4

Stoichiometry

P4 + 5 O2 + 6 H2O 4 H3PO4

What is the mole ratio between O2 and H2O?

What is the mole ratio between H2O and H3PO4?

Stoichiometry

P4 + 5 O2 + 6 H2O 4 H3PO4

What is the mole ratio between O2 and H2O?

5mol O2/ 6mol H2O

What is the mole ratio between H2O and H3PO4?

6mol H2O/ 4mol H3PO4

Moles to Moles

• We use the mole ratio (from the balanced reaction) as a conversion factor

P4 + 5 O2 + 6 H2O 4 H3PO4

• If we started with 6mol of O2, how many moles of H3PO4 would we produce? (Set up the conversion!)

Moles to Moles

P4 + 5 O2 + 6 H2O 4 H3PO4

• If we started with 6mol of O2, how many moles of H3PO4 would we produce? (Set up the conversion!)

6mol O2 x 4mol H3PO4 = 24mol H3PO4 = 4.8mol

1 5mol O2

5mol

4.8 mol of H3PO4

Moles to Moles to Grams

P4 + 5 O2 + 6 H2O 4 H3PO4

• We determined that if we start with 6mol O2, we produce 4.8mol H3PO4. How many grams is this? (Continue with the same conversion set-up! Round molar mass to the nearest whole number.)

Moles to Moles to Grams

P4 + 5 O2 + 6 H2O 4 H3PO4

• We determined that if we start with 6mol O2, we produce 4.8mol H3PO4. How many grams is this?

6mol O2 x 4mol H3PO4 x 98g H3PO4 = 470.4g

1 5mol O2 1mol H3PO4

Grams to Moles to Moles to Grams

P4 + 5 O2 + 6 H2O 4 H3PO4

• If we start with 25g of O2, how many grams of H3PO4 will we produce? (Round molar masses to the nearest whole number)

Grams to Moles to Moles to Grams

P4 + 5 O2 + 6 H2O 4 H3PO4

• If we start with 25g of O2, how many grams of H3PO4 will we produce?

25g O2 x 1mol O2 x 4mol H3PO4 x 98g H3PO4 =

1 32g 5mol O2 1mol H3PO4

=61.25g H3PO4

Exit Ticket #4

Limiting Reactant

• Limiting Reactant: the reactant that limits the amount of product that can be formed in a reaction

• Hint: when the word “excess” is used in a problem, it means there is MORE than enough of that reactant, so that reactant will not be limiting

Finding the Limiting Reactant

• For each of your reactants, multiply the number of MOLES you start with by the coefficient in front of that compound in the balanced reaction

• Whichever compound has the smaller number is the limiting reactant


1 C3H8 + 5 O2 3 CO2 + 4 H2O

You have 3mol C3H8 of and 1.8mol of O2. What is the limiting reactant?


1 C3H8 + 5 O2 3 CO2 + 4 H2O

You have 3mol C3H8 of and 1.8mol of O2. What is the limiting reactant?

C3H8: 3mol x 1 = 3mol Limiting

O2: 1.8mol x 5 = 9mol


1 C3H8 + 5 O2 3 CO2 + 4 H2O

You have 7.6mol C3H8 of and 2.3mol of O2. What is the limiting reactant?


1 C3H8 + 5 O2 3 CO2 + 4 H2O


C3H8: 7.6mol x 1 = 7.6mol Limiting

O2: 2.3mol x 5 = 11.5mol


1 C3H8 + 5 O2 3 CO2 + 4 H2O


C3H8: 15.9mol x 1 = 15.9mol

O2: 3.1mol x 5 = 15.5mol Limiting


1 C3H8 + 5 O2 3 CO2 + 4 H2O


C3H8: 7.5mol x 1 = 7.5mol

O2: 1.5mol x 5 = 7.5mol No limiting reactant


1 C3H8 + 5 O2 3 CO2 + 4 H2O

You have 45g C3H8 of and 40g of O2. What is the limiting reactant?


1 C3H8 + 5 O2 3 CO2 + 4 H2O


*Since we’re given grams, we must convert to moles!


1 C3H8 + 5 O2 3 CO2 + 4 H2O


C3H8: 45g x 1mol = 1.02mol x 1 = 1.02mol (limiting)

44.097g

O2: 40g x 1mol = 1.25mol x 5 = 6.25mol

31.998g

Determining Yield

• Theoretical Yield: the amount of product that a reaction produces if it reaches completion under perfect conditions

– Once you know the limiting reactant, you can calculate how much of your product the reaction should produce

– To set up the conversion, use whatever amount of the limiting reactant was given to you, then set up the conversion from there

Determining Yield

1 C3H8 + 5 O2 3 CO2 + 4 H2O

You have 3mol C3H8 of and 1.8mol of O2. How many grams of H2O will be produced?

C3H8: 3mol x 1 = 3mol Limiting O2: 1.8mol x 5 = 9mol

Determining Yield

1 C3H8 + 5 O2 3 CO2 + 4 H2O

You have 3mol C3H8 of and 1.8mol of O2. How many grams of H2O will be produced?

C3H8: 3mol x 1 = 3mol Limiting O2: 1.8mol x 5 = 9mol

3mol C3H8 x 4mol H2O x 18g H2O = 216g H2O

1 1mol C3H8 1mol H2O

Determining Yield

1 C3H8 + 5 O2 3 CO2 + 4 H2O

You have 7.6mol C3H8 of and 2.3mol of O2. How many grams of CO2 are produced?

C3H8: 7.6mol x 1 = 7.6mol Limiting O2: 2.3mol x 5 =11.5mol

Determining Yield

1 C3H8 + 5 O2 3 CO2 + 4 H2O

You have 7.6mol C3H8 of and 2.3mol of O2. How many grams of CO2 are produced?

C3H8: 7.6mol x 1 = 7.6mol Limiting O2: 2.3mol x 5 =11.5mol

7.6mol C3H8 x 3mol CO2 x 44g CO2 = 1,003g H2O

1 1mol C3H8 1mol CO2

Determining Yield

1 C3H8 + 5 O2 3 CO2 + 4 H2O

You have 200g C3H8 of and 200g of O2. How many grams of H2O are produced?

Determining Yield1 C3H8 + 5 O2 3 CO2 + 4 H2O

You have 200g C3H8 of and 200g of O2. How many grams of H2O are produced?

C3H8: 200g x 1mol = 4.55mol x 1 = 4.55mol: Limiting

44gO2: 200g x 1mol = 6.25mol x 5 = 31.25mol

32g200g C3H8 x 1mol C3H8 x 4mol H2O x 18g H2O = 327.3g

1 44g C3H8 1mol C3H8 1mol H2O

Percent Yield (Recovery)

• Percent Yield (Percent Recovery): the percent of the theoretical yield that was obtained during an experiment– Will always be less than 100%

Percent Yield = Actual Yield x 100%

Theoretical Yield


Previously, we calculated that 216g H2O would be produced if we started with 3mol C3H8 and 1.8mol of O2. If we recovered 200g, what was

our percent yield (percent recovery)?


Previously, we calculated that 216g H2O would be produced if we started with 3mol C3H8 and 1.8mol of O2. If we recovered 200g, what was our percent yield (percent recovery)?

% Yield = 200g x 100 = 92.6% 216g


Previously, we calculated that 1,003g H2O would be produced if we started with 7.6mol C3H8 and 2.3mol of O2. If we recovered 578g, what was our percent yield (percent recovery)?

% Yield = 578g x 100 = 57.6% 1,003g

Summary of Stoichiometry

1. Balance the reaction.2. Find the limiting reactant by multiplying

starting moles by the coefficients in the reaction.

3. Determine the theoretical yield (conversions).

4. If given actual yield, determine percent yield.

Making Solutions: Molarity

• Molarity: the concentration of a solution, measured in moles of solute per liter of solvent

M = mol solute / L solvent

Solute: what you’re dissolvingSolvent: what does the dissolving (usually a

liquid)

Making Solutions

• If you have 10mol of HCl dissolved in 0.75L of water, what is the molarity of the solution?

Making Solutions

• If you have 10mol of HCl dissolved in 0.75L of water, what is the molarity of the solution?

M = mol/ L M = 10mol/ 0.75L = 13.3M

Making Solutions

• If you have a 6M solution of HCl that is 300mL, how many moles of HCl are dissolved?

Making Solutions

• If you have a 6M solution of HCl that is 300mL, how many moles of HCl are dissolved?

M = mol/ L300mL x 1L/ 1,000mL = 0.3L6M = mol/ 0.3L*0.3L = *0.3L 1.8mol

Making Solutions

• You’re trying to make a 1.25M solution of HCl. If you have 500g of HCl, what volume must the solution be in L? In mL?

Making Solutions

• You’re trying to make a 1.25M solution of HCl. If you have 500g of HCl, what volume must the solution be in L? In mL?

M = mol/L500g x 1mol/ 36g = 13.9mol1.25M = 13.9mol/ L11.1L 11.1L x 1,000mL/ 1L = 11, 100L

Dilutions

• Dilution: adding solvent to a solution to decrease the molarity (concentration)

• The product of the initial volume and molarity = the product of the final volume and molarity

M1V1 = M2V2

M = concentration (molarity)

V = volume

Dilutions Example

• You have a 250mL stock solution of 5M HCl. You need to make a dilution to 1.5M HCl. What must the final volume be? How much water will you have to add?

Dilutions Example

• You have a 250mL stock solution of 5M HCl. You need to make a dilution to 1.5M HCl. What must the final volume be? How much water will you have to add?

M1V1 = M2V2

5M*250mL = 1.5MV2

1,250M*mL = 1.5MV2

1.5M 1.5M833.3mL = V2

Molarity of Sugar in Sodas Activity• Your task is to determine the molarity

(concentration) of sugar in various beverages. To do this, you must determine:– The amount of sugar in one serving– The volume of one serving

• We will make the following assumptions:– All sugar in the beverage is fructose: C6H12O6 – There are 30mL per 1 fl. oz.

• When you finish, read the article on sugar and tooth decay, and relate the molarity of sugar that you found to the article.

Exit Ticket #6

Exit Ticket #6Show all work, including the original equation. The molar mass of HCl is

36.5g/mol. Round your answers to the nearest tenth.

1. How many mol of HCl are in a 2L solution with a concentration of 6.5M? (1 point)

2. How many grams of HCl are required to make a 2M, 350mL solution? (2 points)

3. What volume of HCl is needed to create a 3.5M solution from a starting mass of 150g? (3 points)

4. You dilute a 5.5M solution of HCl from 100mL to 250mL. What is the final concentration? (1 point)

5. You leave a 200mL solution of 3M HCl out overnight, and some of the water evaporates. If the final volume is 125mL, what is the final concentration? (2 points)

6. You have a 3M solution of HCl. You add water to the solution until you reach a final concentration of 1.2M. If you started with 175mL, how much water did you add? (3 points)

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