Question

name elfidio mejía date 9/18/25
a. percentage composition
percent composition
determine the percentage composition of sodium carbonate (na₂co₃)?
molar mass percent composition
na = 2(23.00)=46.0 % na $\frac{46.0g}{106g}$×100% = 43.4%
c = 1(12.01)=12.0 % c $\frac{12.0g}{106g}$×100% = 11.3%
o = 3(16.00)=48.0 % o $\frac{48.0g}{106g}$×100% = 45.3%
mm = 106 g
percent composition

• percentage composition of a compound gives the relative amount of each element present.
• % = $\frac{mass element}{mass compound}$×100
• the sum of percentages of each element in a compound should be = 100.

1. determine the percentage composition of h₂o.
2. determine the percentage composition of al₂(co₃)₃.
3. determine the percentage composition of c₆h₁₂o₆.

b. empirical and molecular formula

1. a compound consists of 72.2% magnesium and 27.8% nitrogen by mass. what is the empirical formula?

element	mass (g)	atomic mass	mole	mole ratio

empirical formula: ____

Explanation:

Step1: Calculate molar mass of \(H_2O\)

The atomic mass of \(H = 1.01\ g/mol\) and \(O=16.00\ g/mol\). The molar mass of \(H_2O\), \(MM = 2\times1.01+16.00=18.02\ g/mol\)

Step2: Calculate percentage of \(H\)

\(\%H=\frac{2\times1.01\ g}{18.02\ g}\times 100\% = 11.19\%\)

Step3: Calculate percentage of \(O\)

\(\%O=\frac{16.00\ g}{18.02\ g}\times 100\%=88.81\%\)

Step4: Calculate molar mass of \(Al_2(CO_3)_3\)

Atomic masses: \(Al = 27.00\ g/mol\), \(C = 12.01\ g/mol\), \(O=16.00\ g/mol\). \(MM=2\times27.00 + 3\times12.01+9\times16.00=233.99\ g/mol\)

Step5: Calculate percentage of \(Al\)

\(\%Al=\frac{2\times27.00\ g}{233.99\ g}\times 100\%\approx23.07\%\)

Step6: Calculate percentage of \(C\)

\(\%C=\frac{3\times12.01\ g}{233.99\ g}\times 100\%\approx15.46\%\)

Step7: Calculate percentage of \(O\)

\(\%O=\frac{9\times16.00\ g}{233.99\ g}\times 100\%\approx61.47\%\)

Step8: Calculate molar mass of \(C_6H_{12}O_6\)

Atomic masses: \(C = 12.01\ g/mol\), \(H = 1.01\ g/mol\), \(O = 16.00\ g/mol\). \(MM=6\times12.01 + 12\times1.01+6\times16.00 = 180.18\ g/mol\)

Step9: Calculate percentage of \(C\)

\(\%C=\frac{6\times12.01\ g}{180.18\ g}\times 100\%\approx40.00\%\)

Step10: Calculate percentage of \(H\)

\(\%H=\frac{12\times1.01\ g}{180.18\ g}\times 100\%\approx6.71\%\)

Step11: Calculate percentage of \(O\)

\(\%O=\frac{6\times16.00\ g}{180.18\ g}\times 100\%\approx53.29\%\)

Step12: For the empirical - formula problem of the \(Mg - N\) compound

Assume we have a \(100 - g\) sample. So, mass of \(Mg = 72.2\ g\) and mass of \(N=27.8\ g\)
Atomic mass of \(Mg = 24.31\ g/mol\), atomic mass of \(N = 14.01\ g/mol\)
Moles of \(Mg,n_{Mg}=\frac{72.2\ g}{24.31\ g/mol}\approx2.97\ mol\)
Moles of \(N,n_N=\frac{27.8\ g}{14.01\ g/mol}\approx1.98\ mol\)

Step13: Find the mole - ratio

Divide each number of moles by the smaller number of moles (\(n_N\)). \(\frac{n_{Mg}}{n_N}=\frac{2.97\ mol}{1.98\ mol}\approx1.5\), \(\frac{n_N}{n_N} = 1\)
Multiply by 2 to get whole - number ratio. The mole - ratio of \(Mg:N = 3:2\)
The empirical formula is \(Mg_3N_2\)

Answer:

1. For \(H_2O\): \(\%H\approx11.19\%\), \(\%O\approx88.81\%\)
2. For \(Al_2(CO_3)_3\): \(\%Al\approx23.07\%\), \(\%C\approx15.46\%\), \(\%O\approx61.47\%\)
3. For \(C_6H_{12}O_6\): \(\%C\approx40.00\%\), \(\%H\approx6.71\%\), \(\%O\approx53.29\%\)
4. For the \(Mg - N\) compound, empirical formula: \(Mg_3N_2\)