chemistry quiz

• Went to the chemical store and bought a 4L jug of concentrated hydrochloric acid with conc. Of 18.6M. Plan to  dilute aliquot to produce 1.5 L of a .250 M HCl. How much stock solution should you use?

1: 1500 mL

2. (18.6)(x)=(.250)(1500)

3. solve for x

4. 18.6(x)=375

5. x = 20.2 mL

dilution formula

M₁V₁ = M₂V₂

M₁

molarity of stock

V₁

volume of stock

M₂

molarity of dilute solution

V₂

volume of dilute solution

Adenine is one of the four bases found in DNA. Its percent composition was found to be 44.44% C, 3.73% H, and 51.84% N. A sample containing 0.500 moles of adenine weighs 67.6 grams. What are the empirical and molecular formulas for adenine?

1. Find grams of each element (the percent divided by 100 times total mass)

2. convert grams to moles (grams from step 1 divided by atomic mass of each)

3. find ratio (divide each mole value by whatever was the smallest from step 2)

4. Each WHOLE number is the ratio; in this case, empirical formula is CHN (1:1:1)

5. molecular formula: add up the mass from empirical formula ( C₁ + H₁ + N₁) (in this case, 27.028 g/mol)

6. now find mass of the actual sample, divide your grams given by moles given (in this case, 67.6g / .500 mol = 135.2 g/mol )

7. Divide the number from step 6 by the empirical mass (here, 135.2 / 27.028 = 5.00)

8. So, molecular formula has 5 times the empirical —> C₅H₅N₅

If you dissolve .25 mol of iron (III) chloride in water (FeCl₃) to create .35L solution, whats the molarity of chloride ions

1. find molarity of solution (moles / liters)(here it would be 0.25 / 0.35 = 0.7143 M)

2. find how many chloride ions each formula unit gives (here, it is 3Cl^- so for every 1 mole of FeCl₃ there are 3 moles of Cl^-)

3. find the molarity of chloride ions (3 times the molarity of FeCl₃ which gives 2.14 M)

How many moles of ammonium ions are present in 200 mL of a 1.5 M ammonium phosphate solution?

1. Convert mL to L —> 200 mL = 0.200 L

2. Calculate moles of ammonium phosphate (M times Volume) (1.5 times 0.200 gives us 0.300 moles of (NH₄)₃PO₄

3. find how many ammonium ions per 1 mole of ammonium phosphate (here, there are 3 NH₄ ions)

4. Find moles of ammonium ions (moles from step 2 times ions from step 2)(here it is 0.300 times 3 NH₄ ions, which gives us 0.900 moles of NH₄)

What’s the mass (in kg) of calcium nitrate present in 200.0 L of a 1.556 M solution of calcium nitrate

1. find moles of calcium nitrate (1.556 times 200, which gives us 311.2 mol)

2. Find the molar mass of Ca(NO₃)₂ (1 Ca, 2 N, 6 O add up to give total of 164.10 g/mol)

3. find mass in grams (multiply moles from step 1 by the g/mol from step 2)(here it is 311.2 / 164.10 which equals 51, 072.72 grams)

4. Convert to kg (51,072.2 divided by 1,000 gives us 51.1 kg of Ca(NO₃)₂

If you want to prepare a 0.25 M solution of potassium iodide (166.00 g/mol) using 35 milligrams of potassium iodide, what should be the volume of the solution in milliliters

1. convert milligrams to grams (35 mg = 0.035 g)

2. convert grams to moles (grams given divided by the molar mass given)(here it would be 0.035g / 166.00 g/mol = 2.1084 × 10^-4 mol)

3. use molarity formula to find volume (M = moles / liters but we need to do reverse, so its V = moles / M) (2.1084 × 10^-4 / 0.25 = 8.4 × 10^-4 L)

4. Convert those liters to mL (8.4 × 10^-4 × 1,000 = 8.4 × 10^-1)

You dilute a 4.5 mL aliquot of a 2.3 M stock solution of potassium permanganate to a final volume of 0.50 L. What is the concentration of the diluted solution?

1. get out your dilution formula

2. we are solving for M₂ (molarity/concentration of dilute solution)

3. (2.3)(0.0045) = (X)(0.50)

4. 0.01035 = (X)(0.50)

5. 0.0207 M

A deep red crystalline hydrate compound was found to be 24.77% cobalt, 29.80% chlorine, and 45.43% water by mass. What is the empirical formula for this compound (write it using proper notation for hydrates)?

1. Convert percentage to moles (assume 100g of compound) (24.77g, 29.80g, 45.43g)(divide each of these masses by their periodic table mass)

2. find the ratio by dividing them all by the smallest mole, in this case it was Cobalt with 0.4206, so each molar amount is divided by that number)

3. You get: Co = 1, Cl = 2, H₂O = 6

4. CoCl₂ × 6H₂O