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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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yulyashka [42]

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2 years ago
Suppose that 25.0 mL of 0.10 M CH3COOH (aq) is titrated with 0.10 M NaOH (aq). What is the pH after the addition of 10.0 mL of 0
olya-2409 [2.1K]

Answer:

pH=-1.37

Explanation:

We are given that 25 mL of 0.10 M CH_3COOH is titrated with 0.10 M NaOH(aq).

We have to find the pH of solution

Volume of CH_3COOH=25mL=0.025 L

Volume of NaoH=0.01 L

Volume of solution =25 +10=35 mL=\frac{35}{1000}=0.035 L

Because 1 L=1000 mL

Molarity of NaOH=Concentration OH-=0.10M

Concentration of H+= Molarity of CH_3COOH=0.10 M

Number of moles of H+=Molarity multiply by volume of given acid

Number of moles of H+=0.10\times 0.025=0.0025 moles

Number of moles of OH^-=0.10\times 0.01=0.001mole

Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles

Concentration of H+=\frac{0.0015}{0.035}=4.28\times 10^{-2} m/L

pH=-log [H+]=-log [4.28\times 10^{-2}]=-log4.28+2 log 10=-0.631+2

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6 0
3 years ago
What is the molarity of a solution that contains 1000.0 mg of AgNO3 that has been dissolved in 500 mL of water
Vaselesa [24]

0.012moldm⁻³

Explanation:

Given parameters:

Mass of AgNO₃  = 1000mg

Volume of water = 500mL

Unknown:

Molarity of solution  = ?

Solution:

The molarity of a solution is the number of moles of a solute dissolved in volume of solvent.

 Molarity = \frac{xnumber of moles}{Volume}

 

Number of moles of AgNO₃  = ?

   Number of moles = \frac{mass}{molar mass}

Molar mass of AgNO₃ = 108 + 14 + 3(16) = 170g/mol

   convert mass to g;

      1000mg = 1g

 Number of moles  = \frac{1}{170}  = 0.00588moles

   convert the given volume to dm³;

       1000mL  = 1dm³;

        500mL = 0.5dm³

Now solve;

  Molarity = \frac{0.00588}{0.5}  = 0.012moldm⁻³

learn more:

Molarity brainly.com/question/9324116

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4 0
3 years ago
Determine how many ml of water you need to remove, by evaporation, if you have a 500 ml of 10.20 M HNO3 dilute solution and you
Ludmilka [50]

The total volume of water that would be removed will be 75 mL

<h3>Dilution equation</h3>

Using the dilution equation:

M1V1 = M2V2

In this case, M1 = 500 mL, V1 = 10.20 M, M2 = 12 M

Substitute:

V2 = 500 x 10.20/12

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The final volume in order to arrive at 12 M HNO3 would be 425 mL from the initial 500 mL. Thus, the total amount of water that will be removed by evaporation can be calculated as:

500 - 425 = 75 mL

More on dilution can be found here: brainly.com/question/7208939

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