A sample of paint has a volume of 0.5 L and a density of 5.64 g/mL. The sample is found to
contain 16.6 g Pb3N2(s). Determine the grams of lead (Pb) in the sample.
16.6 * (3Pb)/(3Pb + 2N)
replace the elements by their atomic wts.
what about the density and volume?
To determine the grams of lead (Pb) in the sample, we need to use the given information about the volume and density of the paint along with the mass of the Pb3N2 compound.
First, we should determine the mass of the paint sample. The volume of the paint is given as 0.5 L, and the density is given as 5.64 g/mL. We can use the formula:
mass = volume × density
mass = 0.5 L × 5.64 g/mL
mass = 2.82 g
So, the mass of the paint sample is 2.82 g.
Next, we need to determine the mass of Pb in the Pb3N2 compound. The compound formula tells us that there are 3 Pb atoms for every 2 N atoms. The molar mass of Pb3N2 can be calculated by adding together the atomic masses of three Pb atoms and two N atoms:
3 × atomic mass of Pb + 2 × atomic mass of N = molar mass of Pb3N2
3 × 207.2 g/mol + 2 × 14.0 g/mol = 393.6 g/mol
Therefore, the molar mass of Pb3N2 is 393.6 g/mol.
Now, we can set up a ratio to determine the grams of Pb in the compound:
grams of Pb3N2 : grams of Pb
393.6 g : 3 × atomic mass of Pb (207.2 g) = 16.6 g : x
Solving for x, we get:
x = (16.6 g × 207.2 g) / 393.6 g
x ≈ 8.75 g
Therefore, there are approximately 8.75 grams of lead (Pb) in the sample of paint.