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3 years ago

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The functional group in organic chemistry is the set of __________ attached to the carbon skeleton that often determine the prop

erties of an organic compound.

1 answer:

IrinaK [193]3 years ago

3 0

<span>The functional group in organic chemistry is the set of atoms attached to the carbon skeleton that often determine the properties of an organic compound.

Following are major functional groups that are present in an organic compound. carboxylic acid, ester, ether, amine, amide, thiol, halide, alkene, alkyne, phenol, alcohol, aldehyde, ketone, etc.</span>

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The osmotic pressure of a solution containing 2.04 g of an unknown compound dissolved in 175.0 mLof solution at 25 ∘C is 2.13 at

kherson [118]

<u>Answer:</u> The molecular formula of the compound is $C_4H_{10}O_4$

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 2.13 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Given mass of compound = 2.04 g

Volume of solution = 175.0 mL

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$2.13atm=1\times \frac{2.04\times 1000}{\text{Molar mass of compound}\times 175.0}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\text{Molar mass of compound}=\frac{1\times 2.04\times 1000\times 0.0821\times 298}{2.13\times 175.0}=133.9g/mol$

<u>Calculating the molecular formula:</u>

The chemical equation for the combustion of compound 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 $CO_2=36.26g$

Mass of $H_2O=14.85g$

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

So, in 36.26 g of carbon dioxide, $\frac{12}{44}\times 36.26=9.89g$ of carbon will be contained.

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

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

So, in 14.85 g of water, $\frac{2}{18}\times 14.85=1.65g$ of hydrogen will be contained.

Mass of oxygen in the compound = (22.08) - (9.89 + 1.65) = 10.54 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 = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{9.89g}{12g/mole}=0.824moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.65g}{1g/mole}=1.65moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{10.54g}{16g/mole}=0.659moles$

<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 0.659 moles.

For Carbon = $\frac{0.824}{0.659}=1.25\approx 1$

For Hydrogen = $\frac{1.65}{0.659}=2.5$

For Oxygen = $\frac{0.659}{0.659}=1$

Converting the mole fraction into whole number by multiplying the mole fraction by '2'

Mole fraction of carbon = (1 × 2) = 2

Mole fraction of oxygen = (2.5 × 2) = 5

Mole fraction of hydrogen = (1 × 2) = 2

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

The ratio of C : H : O = 2 : 5 : 2

The empirical formula for the given compound is $C_2H_5O_2$

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 = 133.9 g/mol

Mass of empirical formula = 61 g/mol

Putting values in above equation, we get:

$n=\frac{133.9g/mol}{61g/mol}=2$

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

$C_{(2\times 2)}H_{(5\times 2)}O_{(2\times 2)}=C_4H_{10}O_4$

Hence, the molecular formula of the compound is $C_4H_{10}O_4$

4 0

3 years ago

An experiment shows that a 250 −mL gas sample has a mass of 0.436 g at a pressure of 742 mmHg and a temperature of 27 ∘C.

icang [17]

Answer:

41.9 g/ mol hope that helps you out

Explanation:

d=p.m/ r.t

8 0

3 years ago

Please help me, I really need it, God Bless you <3

gladu [14]

#1

$\\ \sf\longmapsto {}^{88}_{38}Sr^{2+}$

As 2 electrons are donated Hence the No of electrons =38-2=36

#2

$\\ \sf\longmapsto {}^{51}_{23}V^{3+}$

As 3 electrons donated the no of electrons=23-3=20

#3

$\\ \sf\longmapsto {}^{55}_{26}Fe^{3+}$

No of neutrons=55-26=29

#4

$\\ \sf\longmapsto {}^{112}_{48}$

As 2 electrons are donated no of electrons =48-2=46

#5

$\\ \sf\longmapsto {}^{210}_{82}Pb^{2+}$

No of neutrons=210-82=128

7 0

3 years ago

Many elements bond to carbon by exchanging electrons. true or false.

Sliva [168]

If a metal is bonded to carbon, they form ionic bonds. If a nonmetal forms a bond with carbon, they form covalent bond. Ionic bond requires metals to give away electrons to form cations and nonmetal to gain/accept electron to form anions. Covalent bond requires sharing of electrons between both elements. So if you meant exchange as in covalent bond, then no. Not many nonmetal elements form bonds with carbon as there are more metal elements in the periodic table.

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3 years ago

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Define Nuclear Fission with an example of it

ohaa [14]

Answer:

Fission is the splitting of an atomic nucleus into two or more lighter nuclei accompanied by energy release. ... The energy released by nuclear fission is considerable. For example, the fission of one kilogram of uranium releases as much energy as burning around four billion kilograms of coal

Explanation:

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3 years ago

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