Question

Determine the percentage composition of the following compounds: a. \({NaClO}\) c. \(C_{2} {H}_{5} {COOH}\) b. \({H}_{2} {SO}_{3}\) d. \({BeCl}_{2}\)

Short answer

{NaClO}: {Na}: 30.88%, {Cl}: 47.61%, {O}: 21.51%}.{H2SO3}: {H}: 2.46%, {S}: 39.06%, {O}: 58.49%}.{C2H5COOH}: {C}: 48.64%, {H}: 8.16%, {O}: 43.20%}.{BeCl2}: {Be}: 11.28%, {Cl}: 88.72%}.

Step-by-step solution

- Determine the molar mass of each compound

Calculate the molar mass by adding the atomic masses of each element in the compound.For \text{NaClO}:\(M_{\text{Na}} = 22.99 \text{g/mol}\)\(M_{\text{Cl}} = 35.45 \text{g/mol}\)\(M_{\text{O}} = 16.00 \text{g/mol}\)\(M_{\text{NaClO}} = 22.99 + 35.45 + 16.00 = 74.44 \text{g/mol}\)For \text{H}_{2} \text{SO}_{3}:\(2 \times M_{\text{H}} = 2 \times 1.008 = 2.016 \text{g/mol}\)\(M_{\text{S}} = 32.07 \text{g/mol}\)\(3 \times M_{\text{O}} = 3 \times 16.00 = 48.00 \text{g/mol}\)\(M_{\text{H}_{2} \text{SO}_{3}} = 2.016 + 32.07 + 48.00 = 82.086 \text{g/mol}\)For \text{C}_{2} \text{H}_{5} \text{COOH}:\(3 \times M_{\text{C}} = 3 \times 12.01 = 36.03 \text{g/mol}\)\(6 \times M_{\text{H}} = 6 \times 1.008 = 6.048 \text{g/mol}\)\(2 \times M_{\text{O}} = 2 \times 16.00 = 32.00 \text{g/mol}\)\(M_{\text{C}_{2} \text{H}_{5} \text{COOH}} = 36.03 + 6.048 + 32.00 = 74.078 \text{g/mol}\)For \text{BeCl}_{2}:\(M_{\text{Be}} = 9.012 \text{g/mol}\)\(2 \times M_{\text{Cl}} = 2 \times 35.45 = 70.90 \text{g/mol}\)\(M_{\text{BeCl}_{2}} = 9.012 + 70.90 = 79.912 \text{g/mol}\)

- Determine the percentage composition of each element

Use the formula \left(\frac{ \text{mass of the element } }{ \text{molar mass of the compound } } \right) \times 100 \%For \text{NaClO}:\( \text{\text{Na}}: \left(\frac{22.99}{74.44}\right) \times 100 = 30.88\%\)\( \text{\text{Cl}}: \left(\frac{35.45}{74.44}\right) \times 100 = 47.61\%\)\( \text{\text{O}}: \left(\frac{16.00}{74.44}\right) \times 100 = 21.51\%\)For \text{H}_{2} \text{SO}_{3}:\( \text{\text{H}}: \left(\frac{2.016}{82.086}\right) \times 100 = 2.46\%\)\( \text{\text{S}}: \left(\frac{32.07}{82.086}\right) \times 100 = 39.06\%\)\( \text{\text{O}}: \left(\frac{48.00}{82.086}\right) \times 100 = 58.49\%\)For \text{C}_{2} \text{H}_{5} \text{COOH}:\( \text{\text{C}}: \left(\frac{36.03}{74.078}\right) \times 100 = 48.64\%\)\( \text{\text{H}}: \left(\frac{6.048}{74.078}\right) \times 100 = 8.16\%\)\( \text{\text{O}}: \left(\frac{32.00}{74.078}\right) \times 100 = 43.20\%\)For \text{BeCl}_{2}:\( \text{\text{Be}}: \left(\frac{9.012}{79.912}\right) \times 100 = 11.28\%\)\( \text{\text{Cl}}: \left(\frac{70.90}{79.912}\right) \times 100 = 88.72\%\)

Key concepts

Molar Mass Calculation

Understanding molar mass is essential in chemistry. The molar mass of a compound is the sum of the atomic masses of all the atoms in its chemical formula. This value is usually expressed in grams per mole (g/mol). For instance, the molar mass of sodium hypochlorite (NaClO) is calculated by adding the atomic masses of sodium (Na), chlorine (Cl), and oxygen (O).

When computing, you simply add up these atomic masses:

• Molar mass of NaClO: \( M_{\text{NaClO}} = 22.99 \text{ g/mol (Na)} + 35.45 \text{ g/mol (Cl)} + 16.00 \text{ g/mol (O)} = 74.44 \text{ g/mol} \)
• Molar mass of H₂SO₃: \( M_{\text{H}_{2} \text{SO}_{3}} = 1.008 \text{ g/mol (H)} \times 2 + 32.07 \text{ g/mol (S)} + 16.00 \text{ g/mol (O)} \times 3 = 82.086 \text{ g/mol} \)
• Molar mass of C₂H₅COOH: \( M_{\text{C}_{2} \text{H}_{5} \text{COOH}} = 12.01 \text{ g/mol (C)} \times 3 + 1.008 \text{ g/mol (H)} \times 6 + 16.00 \text{ g/mol (O)} \times 2 = 74.078 \text{ g/mol} \)
• Molar mass of BeCl₂: \( M_{\text{BeCl}_{2}} = 9.012 \text{ g/mol (Be)} + 35.45 \text{ g/mol (Cl)} \times 2 = 79.912 \text{ g/mol} \)

This step is the foundation for determining the percentage composition.

Chemical Formulas

Chemical formulas are shorthand representations of compounds, indicating which elements and how many atoms of each element are present. They are crucial for understanding molar mass and elemental composition.

For example:

• **NaClO** represents one sodium (Na), one chlorine (Cl), and one oxygen (O) atom.
• **H₂SO₃** indicates the presence of two hydrogen (H) atoms, one sulfur (S) atom, and three oxygen (O) atoms.
• **C₂H₅COOH** shows that the compound contains three carbon (C) atoms, six hydrogen (H) atoms, and two oxygen (O) atoms.
• **BeCl₂** signifies one beryllium (Be) atom and two chlorine (Cl) atoms.

Understanding these notations is key to calculating molar mass and eventually determining the percentage composition of elements in a compound. Chemical formulas not only convey the types of atoms but also the ratios between them.

Elemental Composition

Elemental composition refers to the percentage by mass of each element in a compound. It's calculated by dividing the mass of each element by the total molar mass of the compound, then multiplying by 100.

The steps to finding the elemental composition are:

• Calculate the molar mass of the compound.
• Determine the mass of each element in one mole of the compound.
• Use the formula: \(\( \frac{ \text{mass of the element} }{ \text{molar mass of the compound} } \)\) \times 100 \ to find the percentage.

For instance, in NaClO, you can find the mass percentages of each component:

• Na: \( \ \frac{22.99 \text{ g/mol (Na)}}{74.44 \text{ g/mol (NaClO)}} \times 100 = 30.88\text{%}\)\
• Cl: \( \ \frac{35.45 \text{ g/mol (Cl)}}{74.44 \text{ g/mol (NaClO)}} \times 100 = 47.61\text{%}\)\
• O: \( \ \frac{16.00 \text{ g/mol (O)}}{74.44 \text{ g/mol (NaClO)}} \times 100 = 21.51\text{%}\)\

Elemental composition helps us understand the distribution of elements within a compound.

Atomic Masses

Atomic masses are the weights of individual atoms, measured in atomic mass units (amu). The atomic mass of an element is approximately equal to the sum of the number of protons and neutrons in its nucleus. This value is essential for various calculations in chemistry.

Common atomic masses you need to know include:

• Sodium (Na): 22.99 amu
• Chlorine (Cl): 35.45 amu
• Oxygen (O): 16.00 amu
• Hydrogen (H): 1.008 amu
• Sulfur (S): 32.07 amu
• Carbon (C): 12.01 amu
• Beryllium (Be): 9.012 amu

These values are used to calculate the molar mass of compounds. For example, in NaClO:

• We use the atomic masses of Na, Cl, and O to find the molar mass: \( M_{\text{NaClO}} = 22.99 \text{ amu (Na)} + 35.45 \text{ amu (Cl)} + 16.00 \text{ amu (O)} \)

Knowing atomic masses allows precise molar mass calculations, essential for further chemical computations.