1) When 5.00g of copper reacts with oxygen, the copper oxide product has a mass

Question

1) When 5.00g of copper reacts with oxygen, the copper oxide product has a mass of 5.62g. What is the empirical formula of this copper oxide? MM Cu= 63.5, O= 16.0

2) What is the molarity of acetic acid, CH3COOH (MM= 60.0) in vinegar if a 5.00mL sample of vinegar requires 22.50mL of a 0.195M NaOH solution to neutralize it?

3) A 20.00mL sample of NaCl solution has a mass of 20.75g. After the NaCl solution is evaporated to dryness the dry salt has a mass of 4.35g. Calculate the following concentration for the NaCl solution, MM NaCl= 58.5

(a) % m/m (or % w/w)

(b) % m/v

(c) Moraity (M)

Explanation / Answer

You want to start by determining a balenced equation for the reaction. The balanced equation for this reaction is:

CH3COOH + KOH --> CH3COO- + K + H2O

From this you know that 1 mole of CH3COOH reacts with 1 mole of KOH.

So the # of moles of KOH that you used will tell you the number of moles of Acetic acid that you had in the 10 mL sample.

To find the moles of KOH:
(0.256mol KOH / L) (1L / 1000 mL) (31.45 mL) = 0.00805 mol KOH

1mole KOH per 1 mol of CH3COOH so you have 0.00805 mol CH3COOH.

to find the molarity, you must divide by the volume in L (of you vinegar sample).

so 10 mL is 0.01 L (work out is 10 mL * (1L/1000mL))

now 0.00805 mol CH3COOH / 0.01 L = 0.805 mol/L CH3COOH. It is 0.805 M.

Notice that all of the units cancel to leave you with only the units you need for your answer in each equation. This is always a good way to check your work. Make sure you double check my math skills! (hope this helps you understand.) Cheers~

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