Note: Respond To All Questions From Questions 1 To 2 Below 1. Make a 30 mL solut

Question

Note: Respond To All Questions From Questions 1 To 2 Below

1. Make a 30 mL solution of 120 mg/mL Cu(11)SO4·5H20 in small beaker. Note that Cu(II)SO4 . 5H20, although not very harmful, is an irritant and thus should be handled with gloves and disposed of in an appropriate waste container what amount of Cu(II)SO4·5H2O do you need to measure to make 30 mLs of Question #8: 120 mg/mL? 2. Make your calculations for how you will create the following solutions. You should make the calculations so that you end up with between 3 and 6 mLs of each concentration since the spectrophotometer requires a minimum of 3 mLs for accurate measurement. You may use any of the solutions that you have already made to create further dilutions if desired, but you must make your plan clear here. Be sure to save enough of your 120 mg/mL (3 mLs) to test it too. 100 mg/mL: 80 mg/m: 75 mg/mL: 60 mg/mL: 50 mg/mL 30 mg/mL: 12 mg/mL:

Explanation / Answer

Question 8.

To make calculations easier, round the atomic mass to the nearest whole number: Copper (Cu) is 64g/mole, sulfur (S) is 32g/mole, oxygen (O) is 16g/mole and hydrogen (H) is 1g/mole.

Add the mass of all the atoms in the chemical formula. Because the formula has only one mole of copper atoms, add 64g only one time. For oxygen, however, determine the total number of moles of atoms in the formula and multiply that number by 16g to get the total mass of oxygen in the compound. The equations are: Cu: 64g x 1 = 64 S: 32g x 1 = 32 O: 16g x 4 = 64 H: 1g x 10 = 10 ("5 H2O" means 10 H and 5 O are involved.) O: 16g x 5 = 80

The total is: 64 + 32 + 64 + 10 + 80 = 250g/mole = gram formula mass of CuSO4*5H2O.

Determining the Number of Moles

Write down the molarity formula. Molarity, or concentration, is equal to the number of moles of solute per liter of solution. Simplified, the formula is M = mol/L.

Plug your desired molarity and volume into the molarity formula. If you want to prepare 1L of a 0.2 M solution, for example, plug those values into the formula to solve for moles this way: M = mol/L and 0.2 M = x mol/1L.

Calculate the number of moles of copper (II) sulfate pentahydrate needed. This operation requires a simple cross-multiplication: x = (0.2 M) (1L) = 0.2 mol.

In this example, you would need 0.2 moles of copper (II) sulfate pentahydrate to make 1L of solution.

Converting Moles to Grams

Write down the mole calculations formula. It can be used to convert moles of a substance to grams of a substance and vice versa. Because the gram formula mass represents the number of grams in 1 mole of a substance, you can obtain the required mass for your solution by multiplying the number of moles by the gram formula mass. Simplified, the formula is: Number of grams = (number of moles) (gram formula mass).

Plug the gram formula mass you calculated previously and the number of moles calculated previously into the mole calculations formula. Using the earlier example, 0.2 moles of copper (II) sulfate pentahydrate are needed: Number of grams = (number of moles) (gram formula mass) Number of grams = (0.2 mol) (250g/mol)

Solve for the number of grams of copper (II) sulfate pentahydrate needed. An example is: (0.2 mol) (250g/mol) = 50g.

In that example, you would need to measure 50g of solid copper (II) sulfate pentahydrate in the laboratory and add water to make 1L of solution.

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