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andrew11 [14]
3 years ago
8

A molecule has the empirical formula C4H6O. If its molecular weight is determined to be about 212 g/mol, what is the most likely

molecular formula?
a. C12H18O3
b. C8H12O3
c. C11H16O4
d. C4H6O
e. C2H3O
Chemistry
2 answers:
Blababa [14]3 years ago
6 0

Answer:

The molecular formula is most likely to be first option, \rm C_{12} H_{18} O_{3}.

Explanation:

Both the molecular formula and the empirical formula of a molecule give the types of atoms in that molecule. However, the molecular formula of a molecule gives the exact number of atoms each molecule. On the other hand, the empirical formula gives only a simplified ratio.

For example, if the empirical formula of the molecule is \rm C_4H_6O, then the molecular formula would be (\rm C_4H_6O)_k, or equivalently \mathrm{C}_{4\, k}\mathrm{H}_{6\, k} \mathrm{O}_{k}, where k is a positive whole number (1,\, 2,\, \dots, etc.) The goal here is to find the value of

Look up the relative atomic mass data on a modern periodic table:

  • C: 12.011.
  • H: 1.008.
  • O: 15.999.

The formula mass of \rm C_4 H_6 O would be

\begin{aligned}&M(\rm C_4 H_6 O) \cr &\approx 4 \times 12.011 + 6 \times 1.008 + 15.999 \cr &= 70.091\;\rm g \cdot mol^{-1}\end{aligned}.

The formula mass of \mathrm{C}_{4\, k}\mathrm{H}_{6\, k} \mathrm{O}_{k} would be

\begin{aligned}&M(\mathrm{C}_{4\, k}\mathrm{H}_{6\, k} \mathrm{O}_{k}) \cr &\approx (4\, k) \times 12.011 + (6\, k) \times 1.008 + k \times 15.999 \cr &= ( 4 \times 12.011 + 6 \times 1.008 + 15.999)\, k \\ &= (70.091\, k)\;\rm g \cdot mol^{-1}\end{aligned}.

On the other hand, since the molecule should have a molecular mass of \rm 212\; g \cdot mol^{-1},

70.091\, k \approx 212.

k \approx 3. (Round to the nearest whole number.)

Hence, the molecular formula would be \rm C_{(4 \times 3)} H_{(6 \times 3)}O_{3}, which simplifies to \rm C_{12} H_{18} O_{3}.

krok68 [10]3 years ago
3 0

Answer:

The molecular formula is C12H18O3

Explanation:

Step 1: Data given

The empirical formula is C4H6O

Molecular weight is 212 g/mol

atomic mass of C = 12 g/mol

atomic mass of H = 1 g/mol

atomic mass of O = 16 g/mol

Step 2: Calculate the molar mass of the empirical formula

Molar mass = 4* 12 + 6*1 +16

Molar mass = 70 g/mol

Step 3: Calculate the molecular formula

We have to multiply the empirical formula by n

n = the molecular weight of the empirical formula / the molecular weight of the molecular formula

n = 70 /212 ≈ 3

We have to multiply the empirical formula by 3

3*(C4H6O- = C12H18O3

The molecular formula is C12H18O3

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Taking into account the reaction stoichiometry, you can observe that:

  • one mole of Ca₃P₂ produces 2 mol of PH₃.
  • the mole ratio between phosphine and calcium phosphide is 2 mol PH₃ over 1 mol Ca₃P₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Ca₃P₂ + 6 H₂O  → 3 Ca(OH)₂ + 2 PH₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Ca₃P₂:1 mole
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  • PH₃: 2 moles

The molar mass of the compounds is:

  • Ca₃P₂: 182 g/mole
  • H₂O: 18 g/mole
  • Ca(OH)₂: 74 g/mole
  • PH₃: 34 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Ca₃P₂: 1 mole ×182 g/mole= 182 grams
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Then, by reaction stoichiometry, you can observe that:

  • one mole of Ca₃P₂ produces 2 mol of PH₃.
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What mass of nitrogen is needed to fill an 855 L tank at STP?
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Answer:

It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.

Explanation:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C or 273.15 °K are used and are reference values for gases.

On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

So, in this case:

  • P= 1 atm
  • V= 855 L
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Replacing:

1 atm* 855 L= n* 0.082 \frac{atm*L}{mol*K} * 273.15 K

Solving:

n=\frac{1 atm* 855 L}{0.082\frac{atm*L}{mol*K} *273.15 K }

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Being the molar mass of nitrogen N2 equal to 28 g / mol, you can apply the following rule of three: if there are 28 grams in 1 mole, how much mass is there in 38.17 moles?

mass=\frac{38.17 moles*28 grams}{1 mole}

mass= 1,068.76 grams

<u><em> It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.</em></u>

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