A 0.1153-gram sample of a pure hydrocarbon was burned in a C-H combustion train to produce 0.3986 gram of CO2and 0.0578 gram of H2O. Determine the masses of C and H in the sample and the percentages of these elements in this hydrocarbon.


Answer 1

Answer: The mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.


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

C_xH_y+O_2\rightarrow CO_2+H_2O

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of CO_2=0.3986g

Mass of H_2O=0.0578g

Mass of sample = 0.1153 g

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

So, in 0.3986 g of carbon dioxide, (12)/(44)* 0.3986=0.1087g of carbon will be contained.

  • For calculating the mass of hydrogen:

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

So, in 0.0578 g of water, (2)/(18)* 0.0578=0.0066g of hydrogen will be contained.

To calculate the percentage composition of a substance in sample, we use the equation:

\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of sample}}* 100      ......(1)

  • For Carbon:

Mass of sample = 0.1153 g

Mass of carbon = 0.1087 g

Putting values in equation 1, we get:

\%\text{ composition of carbon}=(0.1087g)/(0.1153g)* 100=94.27\%

  • For Hydrogen:

Mass of sample = 0.1153 g

Mass of hydrogen = 0.0066 g

Putting values in equation 1, we get:

\%\text{ composition of hydrogen}=(0.0066g)/(0.1153g)* 100=5.72\%

Hence, the mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

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The mass of a water balloon is 2 kilograms. The speed that the water ballon is traveling when it hits the ground is 20 meters/second.What is the total kinetic energy of a water balling that hits the ground after it is dropped from a balcony ?



400 Joules


From the question,

The total kinetic energy of the water balling when hits the ground is given as

K.E = 1/2mv².................. Equation 1

Where K.E = Kinetic Energy of water ballon, m = mass of water balloon, v = velocity of water ballon

Given: m = 2 kilograms, v = 20 meters/second.

Substitute these values into equation 1

K.E = (2×20²)/2

K.E = 2×400/2

K.E = 400 Joules

What is the covalent bond for CO?​



Covalent bond or common bond is one of the types of chemical bonds. This connection arises from electronic participation. In fact, atoms that need to receive electrons to achieve stable electron arrangement (noble gas electron arrangement or octagonal arrangement) share electrons in their valence layer with other atoms. In this case, the transfer of electrons from one atom to another does not take place, but only a pair of electrons, called a bonded or shared electron pair, belongs to the nucleus of two atoms.

How many electrons does a Bromine (Br) atom have?(Given Information)
Atomic Number : 35
Neutrons: 45
Charge; -1


35 yooooooooo...........

What is the value for the kinetic energyfor a n = 2 Bohr orbit electron in Joules?



K.E. = 5.4362 × 10⁻¹⁹ J


The expression for Bohr velocity is:

v=(Ze^2)/(2 \epsilon_0* n* h)

Applying values for hydrogen atom,  

Z = 1

Mass of the electron (m_e) is 9.1093×10⁻³¹ kg

Charge of electron (e) is 1.60217662 × 10⁻¹⁹ C

\epsilon_0 = 8.854×10⁻¹² C² N⁻¹ m⁻²

h is Plank's constant having value = 6.626×10⁻³⁴ m² kg / s

We get that:

v=\frac {2.185* 10^6}{n}\ m/s

Given, n = 2


v=\frac {2.185* 10^6}{2}\ m/s

v=1.0925* 10^6\ m/s

Kinetic energy is:

K.E.=\frac {1}{2}* mv^2


K.E.=\frac {1}{2}* 9.1093* 10^(-31)* ({1.0925* 10^6})^2

K.E. = 5.4362 × 10⁻¹⁹ J

Exactly 1.5 g of a fuel burns under conditions of constant pressure and then again under conditions of constant volume. In measurement A the reaction produces 25.9 kJ of heat , and in measurement B the reaction produces 23.3 kJ of heat. Which measurement (A or B) corresponds to conditions of constant pressure? Which one corresponds to conditions of constant volume? Explain.



Process B : constant pressure condition

Process A : constant volume condition


In case of constant pressure, some of the energy is used to do work on the surrounding to keep pressure constant. Due to this, the total heat energy is less than in case of constant volume. In Case of constant Volume all of heat is available, produced in reaction because no work is done.

If we look at our data,we will find that process B has energy 23.3 KJ which is less than process A, the energy of which is 25.9 KJ. It means Process B is occurred at constant pressure condition and Process A has occurred at constant volume condition

A buffer solution contains 0.11 mol of acetic acid and 0.13 mol of sodium acetate in 1.00 L. What is the pH of this buffer?What is the pH of the buffer after the addition of 2?



Henderson Hasselbalch equation: pH = pKa + log [salt]/[acid]

You need to know the pKa for acetic acid. Looking it up one finds it to be 4.76

(a). pH = 4.76 + log [0.13]/[0.10]

= 4.76 + 0.11

= 4.87

(b) KOH + CH3COOH =>H2O + CH3COOK so (acid)goes down and (salt)goes up. Assuming no change in volume, you have 0.10 mol acid - 0.02 mol = 0.08 mol acid and 0.13 mol salt + 0.02 mol = 0.15 mol salt

pH = 4.76 + log [0.15]/[0.08]

= 4.76 + 0.27

= 5.03

Final answer:

The pH of the buffer with 0.11 mol acetic acid and 0.13 mol sodium acetate in 1.00 L is 4.91, calculated using the Henderson-Hasselbalch equation. The second part of the question regarding the pH change after the addition of '2' is unanswerable without further information on what is being added.


To answer the question of what the pH of the buffer solution containing 0.11 mol of acetic acid and 0.13 mol of sodium acetate in 1.00 L is, we can apply the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Where [A-] is the concentration of the acetate ion and [HA] is the concentration of acetic acid. For acetic acid, the pKa is approximately 4.76. Since we have 0.11 mol of acetic acid and 0.13 mol of sodium acetate in 1.00 L solution, the concentrations are 0.11 M and 0.13 M respectively.

Substituting these values into the Henderson-Hasselbalch equation gives:

pH = 4.76 + log(0.13/0.11)

Calculating the log(0.13/0.11) yields approximately 0.15. Therefore:

pH = 4.76 + 0.15 = 4.91

The question "What is the pH of the buffer after the addition of 2?" seems to be incomplete, as it does not specify what '2' refers to. If '2' refers to adding 2 moles of a strong acid or base, for instance, the pH would change significantly and the buffer capacity might be exceeded. The exact effect on pH would depend on the nature of the substance added (acid or base) and its quantity. Without specifics, this part of the question cannot be accurately answered.

The concept of buffer capacity is relevant to discuss here. Buffer capacity refers to the amount of acid or base a buffer can absorb without a significant change in pH. Buffer solutions with higher molar concentrations of both the acid and the corresponding salt will have greater buffer capacity.

Learn more about Buffer Solution pH here: