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Sagot :
Answer:
[tex]\boxed {\boxed {\sf 5.56 \times 10^{23} \ molecules \ CH_4}}[/tex]
Explanation:
We are asked to find how many molecules of methane are in 14.8 grams of the substance.
1. Convert Grams to Moles
First, we convert grams to moles. We use the molar mass, or the mass of 1 mole of a substance. These values are equivalent to the atomic masses found on the Periodic Table, however the units are grams per mole instead of atomic mass units.
We are given the compound methane, or CH₄. Look up the molar mass of the individual elements (carbon and hydrogen).
- C: 12.011 g/mol
- H: 1.008 g/mol
Check the formula for subscripts. Hydrogen (H) has a subscript of 4, so there are 4 moles of hydrogen in 1 mole of methane. We must multiply hydrogen's molar mass by 4, then add carbon's molar mass.
- H₄: 1.008 * 4 = 4.032 g/mol
- CH₄: 12.011 + 4.032 = 16.043 g/mol
Now we use dimensional analysis to convert. To do this, we set up a ratio using the molar mass.
[tex]\frac {16.043 \ g \ CH_4 }{ 1 \ mol \ CH_4}[/tex]
Since we are converting 14.8 grams of methane to moles, we multiply by this value.
[tex]14.8 \ g \ CH_4 *\frac {16.043 \ g \ CH_4 }{ 1 \ mol \ CH_4}[/tex]
Flip the ratio so the units of grams of methane cancel.
[tex]14.8 \ g \ CH_4 *\frac{ 1 \ mol \ CH_4} {16.043 \ g \ CH_4 }[/tex]
[tex]14.8 *\frac{ 1 \ mol \ CH_4} {16.043}[/tex]
[tex]\frac {14.8}{16.043} \ mol \ CH_4= 0.9225207256 \ mol \ CH_4[/tex]
2. Moles to Molecules
Next, we convert moles to molecules. We use Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this problem, the particles are moles of methane. Set up another ratio using Avogadro's Number.
[tex]\frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 \ mol \ CH_4}[/tex]
Multiply by the number of moles we calculated.
[tex]0.9225207256\ mol \ CH_4 * \frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 \ mol \ CH_4}[/tex]
The units of moles of methane cancel.
[tex]0.9225207256* \frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 }[/tex]
[tex]5.55541981 \times 10^{23} \ molecules \ CH_4[/tex]
3. Round
The original measurement of grams (14.8) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 5 in the thousandth place tells us to round the 5 in the hundredth place up to a 6.
[tex]5.56 \times 10^{23} \ molecules \ CH_4[/tex]
14.8 grams of methane is equal to approximately 5.56 × 10²³ molecules of methane.
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