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What information do we need to determine the molecular formula of a compound from the empirical formula?

defon

Answer:

Molecular mass

Explanation:

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Thus if the molecular mass is known, then we can find the value of n which results to molecular formula.

6 0

3 years ago

For the following reaction, KP = 0.455 at 945°C: At equilibrium, is 1.78 atm. What is the equilibrium partial pressure of CH4 in

Alborosie

Answer:

See explanation below

Explanation:

The question is incomplete. However, here's the missing part of the question:

"For the following reaction, Kp = 0.455 at 945 °C:

C(s) + 2H2(g) <--> CH4(g).

At equilibrium the partial pressure of H2 is 1.78 atm. What is the equilibrium partial pressure of CH4(g)?"

With these question, and knowing the value of equilibrium of this reaction we can calculate the partial pressure of CH4.

The expression of Kp for this reaction is:

Kp = PpCH4 / (PpH2)²

We know the value of Kp and pressure of hydrogen, so, let's solve for CH4:

PpCH4 = Kp * PpH2²

*: You should note that we don't use Carbon here, because it's solid, and solids and liquids do not contribute in the expression of equilibrium, mainly because their concentration is constant and near to 1.

Now solving for PpCH4:

PpCH4 = 0.455 * (1.78)²

PpCH4 = 1.44 atm

6 0

3 years ago

A sample of a compound containing boron (B) and hydrogen (H) contains 5.443 g of B and 1.522 g of H. The molar mass of the compo

Svetach [21]

You are calculating the empirical formula of this chemical compound, which is the question with moles, molar mass, and number of moles.

first you divide the mass of BORON by its molar mass(relative formula mass)because there is a formula about moles state: number of moles=mass/molar mass.

So, 5.443/11 is about 0.5. Then, the RFM of H is 1, so the number of mole is 1.522/1=1.522.

Next, you get the number of moles in order is; 0.5 and 1.522. Now we need to look at the ratio between these numbers. 0.5 is smaller so we use it as the ratio of 1. next use 1.522/0.5 is 3.044 which has a greatest common factor of 3. so the empirical formula is BH3.

Now we are going to solve the molecular formula.

the molar mass ofthis compound is 30g, so we're going to find the RFM of the empirical formulsof BH3 first.

11+3=14.

now we see how many times 14 goes into 30. 30/14=2.14 which is about 2.

So now we need to times the subscript of the empirical formula by two.

thus, the molecular formula is B2H6.

To solve this kind of questions, there are many steps:Know what you are calculating about, it's about the molecular formula, so you need to find out the number of moles of each elements. then use the molar mass of the whole compound to calculate the molecular formula.

1) Find the RFM of the element, because that is the molar mass(mass of 1 mole) of this element.

2) number of moles= mass/molar mass. use this formula to help you get the number of moles of each element in this compound

3) look at the relationship between the number of moles of each elements. find out the ratio between them.

4) then use the molarmass of the whole compound to find the molecular formula. molar mass of the whole compound/RFM(molar mass) of the empirical formula of elements= the number you need to multiply by the subscript of the empirical formula to get the molecular formula.

please tell me if i got anything wrong;)

4 0

3 years ago

DATA:

WITCHER [35]

Answer:

Explanation:

trial 1

5 0

3 years ago

how would you write equation to show how buffers H2CO3 and NaHCO3 behave when (a) HCl is added and (b) NaOH is added.

Vlad1618 [11]

H2CO3 <---> H+ + HCO3-
NaHCO3 <---> Na+ + HCO3-

When acid is added in the buffer, the excess H+ of that acid reacts with HCO3- to form H2CO3, and due to this NaHCO3 dissociates into HCO3- to attain the equilibrium. and hence there is no net effect of H+ due to pH remain almost constant. when a base is added to the buffer, the OH- ion of base react eith H+ ion present in buffer, then to attain equilibrium of H+ ion, the H2CO3 dissociates to produce H+ ion, but now there is the excess of HCO3- due to which Na+ ion react with them to attain equilibrium of HCO3-. hence there is again no net change in H+ ion due to which pH remain constant.....

8 0

3 years ago