Question

1. suppose that you did a procedure similar to this lab but using a sample of nickel chloride. you weighed a sample of the original nickel chloride compound, then reacted it to produce nickel metal (ni), which was collected on filter paper and carefully dried. use the data in the table and show your work clearly for the following calculations.

a. mass of nickel formed:
b. moles of nickel formed:
c. mass of chlorine that must have been present in the original sample:
d. moles of chlorine that must have been present in the original sample:
e. empirical formula of the compound (start with the unrounded values from above):

mass of nickel chloride	3.77 g
mass of dry filter paper	0.29 g
mass of filter paper and nickel after complete drying	1.63 g

Explanation:

Step1: Calculate mass of nickel

Subtract mass of dry filter - paper from mass of filter - paper and nickel after drying.
$m_{Ni}=1.63\ g - 0.29\ g=1.34\ g$

Step2: Calculate moles of nickel

Use the molar - mass formula $n=\frac{m}{M}$, where the molar mass of nickel $M_{Ni}=58.69\ g/mol$.
$n_{Ni}=\frac{1.34\ g}{58.69\ g/mol}\approx0.0228\ mol$

Step3: Calculate mass of chlorine

Subtract mass of nickel from mass of nickel chloride.
$m_{Cl}=3.77\ g - 1.34\ g = 2.43\ g$

Step4: Calculate moles of chlorine

Use the molar - mass formula $n=\frac{m}{M}$, where the molar mass of chlorine $M_{Cl}=35.45\ g/mol$.
$n_{Cl}=\frac{2.43\ g}{35.45\ g/mol}\approx0.0686\ mol$

Step5: Determine empirical formula

Find the ratio of moles of nickel to moles of chlorine. Divide each number of moles by the smaller number of moles ($n_{Ni}=0.0228\ mol$).
$\frac{n_{Ni}}{n_{Ni}} = 1$, $\frac{n_{Cl}}{n_{Ni}}=\frac{0.0686\ mol}{0.0228\ mol}\approx3$
The empirical formula is $NiCl_3$.

Answer:

a. $1.34\ g$
b. $0.0228\ mol$
c. $2.43\ g$
d. $0.0686\ mol$
e. $NiCl_3$