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RUDIKE [14]
2 years ago
3

Ascorbic acid (vitamin C) contains C, H, and O. In one combustion analysis, 5.24 g of ascorbic acid yields 7.86 g CO2 and 2.14 g

H2O. Calculate the empirical formula and molecular formula of ascorbic acid given that its molar mass is about 176 g.
Chemistry
1 answer:
prohojiy [21]2 years ago
5 0

Answer : The molecular formula for the given organic compound is C_6H_6O_6

Explanation :

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:

Mass of CO_2=7.86g

Mass of H_2O=2.14g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 7.86 g of carbon dioxide, \frac{12}{44}\times 7.86=2.14g of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 2.14 g of water, \frac{2}{18}\times 2.14=0.238 of hydrogen will be contained.

Mass of oxygen in the compound = (5.24) - (2.14 + 0.238) = 2.86 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.14g}{12g/mole}=0.178moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.238g}{1g/mole}=0.238moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{2.86g}{16g/mole}=0.178moles

Step 2: 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 0.178 moles.

For Carbon = \frac{0.178}{0.178}=1

For Hydrogen  = \frac{0.238}{0.178}=1.33\approx 1

For Oxygen  = \frac{0.178}{0.178}=1

Step 3: Taking the mole ratio as their subscripts.

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

The empirical formula for the given compound is C_1H_1O_1=CHO

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 176 g/mol

Mass of empirical formula = 29 g/mol

Putting values in above equation, we get:

n=\frac{176}{29}=6

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 6)}H_{(1\times 6)}O_{(1\times 6)}=C_6H_6O_6

Thus, the molecular formula for the given organic compound is C_6H_6O_6

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enyata [817]

Answer:

b. 295 pm

Explanation:

To answer this question we need to use the equation of a face-centered cubic laticce:

Edge length = √8 R

<em>Where R is radius of the atom.</em>

<em />

Replacing:

417pm = √8 R

R = 147.4pm is the radius of the atom

As diameter = 2 radius.

Diameter of the metal atom is:

147.4pm* 2 =

295pm

Right answer is:

<h3>b. 295 pm </h3>

8 0
2 years ago
Water is sometimes removed from the products of a reaction by placing the products in a closed container with excess P4O10. Wate
Ganezh [65]

Answer : The correct answer for mole ratio of H₂O : H₃PO₄ = 6: 4 .

Mole ratio :

It is defined as mole of one substance to another substance in a balanced reaction . In the balanced reaction , the coefficient written before the substances are taken as moles of that substance.

The given balanced reaction between P₄H₁₀ and H₂O is :

P₄O₁₀ + 6 H₂O → 4 H₃PO₄

Mole of H₂O = 6

Mole of H₃PO₄ = 4

Hence mole ratio of Water : H₃PO₄ = 6 : 4


8 0
2 years ago
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Based on the results of this lab, write a short paragraph that summarizes how to distinguish physical changes from chemical chan
Kaylis [27]

Physical changes occur when the properties of a substance are retained and/or the materials can be recovered after the change. Chemical changes involve the formation of a new substance. Formation of a gas, solid, light, or heat are possible evidence of chemical change.

6 0
2 years ago
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A vessel contained N2, Ar, He, and Ne. The total pressure in the vessel was 987 torr. The partial pressures of nitrogen, argon,
xz_007 [3.2K]

Answer:

The partial pressure of neon in the vessel was 239 torr.

Explanation:

In all cases involving gas mixtures, the total gas pressure is related to the partial pressures, that is, the pressures of the individual gaseous components of the mixture. Put simply, the partial pressure of a gas is the pressure it exerts on a mixture of gases.

Dalton's law states that the total pressure of a mixture of gases is equal to the sum of the pressures that each gas would exert if it were alone. Then:

PT= P1 + P2 + P3 + P4…+ Pn

where n is the amount of gases present in the mixture.

In this case:

PT=PN₂ + PAr + PHe + PNe

where:

  • PT= 987 torr
  • PN₂= 44 torr
  • PAr= 486 torr
  • PHe= 218 torr
  • PNe= ?

Replacing:

987 torr= 44 torr + 486 torr + 218 torr + PNe

Solving:

987 torr= 748 torr + PNe

PNe= 987 torr - 748 torr

PNe= 239 torr

<u><em>The partial pressure of neon in the vessel was 239 torr.</em></u>

4 0
2 years ago
What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig
Mashutka [201]

Answer:

P_H =2.86

c=1.4\times 10^{-4}

Explanation:

first write the equilibrium equaion ,

C_3H_6O_3  ⇄ C_3H_5O_3^{-}  +H^{+}

assuming degree of dissociation \alpha =1/10;

and initial concentraion of C_3H_6O_3 =c;

At equlibrium ;

concentration of C_3H_6O_3 = c-c\alpha

[C_3H_5O_3^{-}  ]= c\alpha

[H^{+}] = c\alpha

K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}

\alpha is very small so 1-\alpha can be neglected

and equation is;

K_a = {c\alpha \times \alpha}

[H^{+}] = c\alpha = \frac{K_a}{\alpha}

P_H =- log[H^{+} ]

P_H =-logK_a + log\alpha

K_a =1.38\times10^{-4}

\alpha = \frac{1}{10}

P_H= 3.86-1

P_H =2.86

composiion ;

c=\frac{1}{\alpha} \times [H^{+}]

[H^{+}] =antilog(-P_H)

[H^{+} ] =0.0014

c=0.0014\times \frac{1}{10}

c=1.4\times 10^{-4}

6 0
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