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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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<h3>How do we calculate the formula of hydrate?</h3>

The number of moles of water per mole of anhydrous solid (x) will be computed by dividing the number of moles of water by the number of moles of anhydrous solid (x) to find the hydrate's formula.

Moles will be calculated as:
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So formula of hydrate is MgSO₃.6H₂O.

Hence required formula of hydrate compound is MgSO₃.6H₂O.

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<h3>What is equilibrium state?</h3>

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From the equilibrium state reaction will move only that side which will contribute to maintain the stable state. In the forward reaction heat is released as mention in the question. So, when the temperature of reaction is increased then it shifts towards the left side by absorbing the heat and maintain the stability.

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Let's plug in the values in the formula:

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5 0
3 years ago
A chemical engineer is developing a process for producing a new chemical. One step in the process involves allowing a solution o
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Answer:

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Explanation:

<em>The option that would likely increase the rate of reaction would be to use a more concentrated potassium hydroxide.</em>

<u>The concentration of reactants is one of the factors that affect the rate of reaction. The more the concentration of the reactants, the faster the rate of reaction. </u>

Granted that there are enough of the other reactants, increasing the concentration of one of the reactants will lead to an increased rate of reaction.

Hence, using a more concentrated potassium hydroxide which happens to be one of the reactants would likely increase the rate of reaction.

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