What mass of metal is there in a #500*g# mass of #"calcium nitrate"#?

Question

Luna Chen · Accepted Answer

Approx. #120*g# with respect to calcium............

We determine the molar quantity of #"calcium nitrate"#, i.e. we take the quotient #"mass"/"molar mass"#, and get an answer in moles.... And thus.... #(500*cancelg)/(164.09*cancelg*mol^-1)=3.05*1/(mol^-1)=3.05*1/(1/(mol))=3.05*mol# Now given the formula #Ca(NO_3)_2# there are #3.05*mol# with respect to calcium, #6.10*mol# with respect to nitrogen, and #18.3*mol# with respect to oxygen....and the masses of these atoms sum to approx. #500*g#... And thus mass of calcium present in the sample is given by the product...#3.05*molxx40.1*g*mol^-1=??*g#

Wyatt Adkins · Answer

undefined To determine the mass of metal in a given mass of calcium nitrate ($Ca(NO_3)_2$), we need to first find the molar mass of calcium nitrate and then identify the portion of that mass attributed to the metal (calcium).

The molar mass of calcium nitrate can be calculated by summing the atomic masses of its constituent elements: calcium (Ca), nitrogen (N), and oxygen (O).

- Calcium (Ca) has an atomic mass of approximately 40.08 g/mol.
- Nitrogen (N) has an atomic mass of approximately 14.01 g/mol.
- Oxygen (O) has an atomic mass of approximately 16.00 g/mol, but there are three oxygen atoms in calcium nitrate.

Using these values, we can calculate the molar mass of calcium nitrate:

$ \text{Molar mass of Ca(NO}_3\text{)}_2 = (1 \times \text{atomic mass of Ca}) + (2 \times \text{atomic mass of N}) + (6 \times \text{atomic mass of O}) $

$ = (1 \times 40.08\, \text{g/mol}) + (2 \times 14.01\, \text{g/mol}) + (6 \times 16.00\, \text{g/mol}) $

$ = 40.08\, \text{g/mol} + 28.02\, \text{g/mol} + 96.00\, \text{g/mol} $

$ = 164.10\, \text{g/mol} $

Now, to find the mass of metal (calcium) in a 500 g mass of calcium nitrate, we can use the molar mass ratio:

$ \text{Mass of metal (calcium)} = \frac{\text{Molar mass of calcium}}{\text{Molar mass of calcium nitrate}} \times \text{Mass of calcium nitrate} $

$ = \frac{40.08\, \text{g/mol}}{164.10\, \text{g/mol}} \times 500\, \text{g} $

$ \approx 121.96\, \text{g} $

So, there are approximately 121.96 grams of metal (calcium) in a 500 gram mass of calcium nitrate.