The way you calculate the empirical formula is to firstly assume 100g. To find each elements moles you take each elements percentage listed, times it by one mole and divide it by its atomic mass. (ex: moles of K =55.3g x 1 mole/39.1g, therefore there is 1.41432225 moles of Potassium) Once you’ve completed this for every element you list each elements symbol beside it’s number of moles and divide by the smallest number because it can only go into its self once. After you’ve done this, you’ve found your empirical formula, which is the simplest whole number ratio of atoms in a compound. I’ve added an example of a empirical question I completed last semester :)
Answer:
answer
Explanation:
Area : Square Meters,
Volume : Cubic Meters,
Density : Kilograms per Meter Cube,
Velocity : Meters per Second Squared,
Force : Newtons (Mass multiplied by Acceleration or Mass multiplied by Displacement divided by Time squared)…..
Answer:
0.1113 mol
Explanation:
Data Given:
no. of atoms of CH₄= 6.70 x 10²² atoms
no. of moles of methane (CH₄) = ?
Solution:
we will find no. of moles of methane (CH₄)
Formula used
no. of moles = no. of atoms / Avogadro's number
Where
Avogadro's number = 6.022 x 10²³
Put values in above equation
no. of moles = 6.70 x 10²² atoms / 6.022 x 10²³ (atoms/mol)
no. of moles = 0.1113 mol
So,
There are 0.1113 moles of methane.
Answer:
The enthalpy of the nitrogen-nitrogen bond in N2H4 is 162.6 kJ
Explanation:
For the reaction: N2H4(g)+H2(g)→2NH3(g), the enthalpy change of reaction is
ΔH rxn = 2 ΔHºf NH3 - ΔHºf N2H4
but we also know that the ΔH rxn is calculated by accounting the sum of number of bonds formed and bonds broken as follows:
ΔH rxn = 6H (N-H) + 4 (N-H) + 2H (H-H)
where H is the bond enthalpy .When bonds are broken H is positive, and negative when formed, in the product there are 6 N-H bonds , and in the reactants 4 N-H and 1 H-H bonds).
Consulting an appropiate reference handbook or table the following values are used:
ΔHºf (NH3) = -46 kJ/mol
ΔHºf (N2H4) = 95.94 kJ/mol
(The enthalpy of fomation of hydrogen in its standard state is zero)
H (N-H) = 391 kJ
H (H-H) = 432 kJ
H (N-N) = ?
So plugging our values:
ΔH rxn = 2mol ( -46.0 kJ/mol) - 1mol(95.4 kJ/mol) = -187.40 kJ
-187.40 kJ = 6(-391 kJ) + 4 (391 kJ) + 432 + H(N-N)
-187.40 kJ = -350 kJ + H(N-N)
H(N-N) = 162.6 kJ