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Irina-Kira [14]
3 years ago
8

The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of

2.78 mg of ethyl butyrate produces 6.32 mg of CO2 and 2.58 mg of H2O. What is the empirical formula of the compound?
Chemistry
1 answer:
gavmur [86]3 years ago
3 0

<u>Answer:</u> The empirical formula for the given compound is C_3H_6O

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Conversion factor:  1 g = 1000 mg

Mass of CO_2=6.32mg=0.00632g

Mass of H_2O=2.58g=0.00258g

Mass of compound = 2.78 mg = 0.00278 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.00632 g of carbon dioxide, \frac{12}{44}\times 0.00632=0.00172g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.00258 g of water, \frac{2}{18}\times 0.00258=0.000286g of hydrogen will be contained.

  • Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 4.83\times 10^{-5}mol

For Carbon = \frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3

For Hydrogen  = \frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6

For Oxygen  = \frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is C_3H_{6}O_1=C_3H_6O

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The symbol of the period five element that is a member of the pnicitides family
Zepler [3.9K]

The symbol of the period five element that is a member of the pnicitides family are antimony.

Explanation:

Pnictogen family

  • In periodic table, column 15 elements are Pnitogen family.
  • The pnictogen elements are Nitrogen-N, arsenic-As, phosphorus-P, bismuth-Bi, antimony-Sb, ununpentium-Uup.    
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Antimony  

  • Antimony is the element found in period 5 and block p.
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3 years ago
The solubility of magnesium phosphate at a given temperature is 0.173 g/L. Calculate the Ksp at this temperature. After you calc
jek_recluse [69]

Answer: K_{sp}=1.25\times 10^{-14}

pK_{sp}=13.90

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as K_{sp}  

The equation for the ionization of magnesium phosphate is given as:

Mg_3(PO_4)_2\rightarrow 3Mg^{2+}+2PO_4^{3-}

 When the solubility of Mg_3(PO_4)_2 is S moles/liter, then the solubility of Mg^{2+} will be 3S moles\liter and solubility of PO_4^{3-} will be 2S moles/liter.

Thus S = 0.173 g/L or \frac{0.173g/L}{262.8g/mol}=0.00065mol/L

K_{sp}=(3S)^3\times (2S)^2

K_{sp}=108S^5

K_{sp}=108\times (0.00065)^5=1.25\times 10^{-14}

pK_{sp}=-log(K_{sp})=\log (1.25\times 10^{-14})=13.90

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