Question

calculate the mass in grams of each of the following:
a) 1.38 moles of n
b) 2.57 x 10^8 moles of s
c) 1.20 x 10^15 atoms o

calculate the number of atoms present in each of the following
d) 2 moles fe
e) 40.1g ca
f) 4.5g be
g) 184.2g cu

Explanation:

Step1: Recall molar - mass formula

Mass (m) = number of moles (n)×molar mass (M). Molar mass of N is 14.01 g/mol, S is 32.07 g/mol, O is 16.00 g/mol, Fe is 55.85 g/mol, Ca is 40.08 g/mol, Be is 9.01 g/mol, Cu is 63.55 g/mol. Avogadro's number \(N_A=6.022\times 10^{23}\text{ atoms/mol}\).

Step2: Calculate mass for part A

For 1.38 moles of N:
\[m = n\times M=1.38\text{ mol}\times14.01\text{ g/mol}=19.3338\text{ g}\]

Step3: Calculate mass for part B

For \(2.57\times 10^{8}\) moles of S:
\[m = n\times M = 2.57\times 10^{8}\text{ mol}\times32.07\text{ g/mol}=8.24199\times 10^{9}\text{ g}\]

Step4: Calculate moles for part C

First, find moles of O. \(n=\frac{\text{number of atoms}}{N_A}\), so \(n=\frac{1.20\times 10^{15}\text{ atoms}}{6.022\times 10^{23}\text{ atoms/mol}}=1.993\times 10^{-9}\text{ mol}\). Then mass \(m = n\times M=1.993\times 10^{-9}\text{ mol}\times16.00\text{ g/mol}=3.1888\times 10^{-8}\text{ g}\)

Step5: Calculate number of atoms for part D

Number of atoms of Fe: \(N = n\times N_A=2\text{ mol}\times6.022\times 10^{23}\text{ atoms/mol}=1.2044\times 10^{24}\text{ atoms}\)

Step6: Calculate number of moles for part E

\(n=\frac{m}{M}\), for Ca, \(n=\frac{40.1\text{ g}}{40.08\text{ g/mol}} = 1.0005\text{ mol}\). Number of atoms \(N=n\times N_A=1.0005\text{ mol}\times6.022\times 10^{23}\text{ atoms/mol}=6.025\times 10^{23}\text{ atoms}\)

Step7: Calculate number of moles for part F

For Be, \(n=\frac{m}{M}=\frac{4.5\text{ g}}{9.01\text{ g/mol}} = 0.4994\text{ mol}\). Number of atoms \(N=n\times N_A=0.4994\text{ mol}\times6.022\times 10^{23}\text{ atoms/mol}=3.007\times 10^{23}\text{ atoms}\)

Step8: Calculate number of moles for part G

For Cu, \(n=\frac{m}{M}=\frac{184.2\text{ g}}{63.55\text{ g/mol}} = 2.90\text{ mol}\). Number of atoms \(N=n\times N_A=2.90\text{ mol}\times6.022\times 10^{23}\text{ atoms/mol}=1.7464\times 10^{24}\text{ atoms}\)

Answer:

A. 19.33 g
B. \(8.24\times 10^{9}\text{ g}\)
C. \(3.19\times 10^{-8}\text{ g}\)
D. \(1.20\times 10^{24}\text{ atoms}\)
E. \(6.03\times 10^{23}\text{ atoms}\)
F. \(3.01\times 10^{23}\text{ atoms}\)
G. \(1.75\times 10^{24}\text{ atoms}\)