Answer:
Empirical formula: CH₃O
Empirical formula mass = 31 g/mol
Explanation:
Data Given:
Molecular Formula = C₁₀H₃₀O₁₀
Empirical Formula = ?
Empirical Formula mass =
Solution
Empirical Formula:
Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.
So,
The ratio of the molecular formula should be divided by whole number to get the simplest ratio of molecule
As
C₁₀H₃₀O₁₀ Consist of 10 Carbon (C) atoms, 30 Hydrogen (H) atoms, and 10 Oxygen (O) atoms.
Now
Look at the ratio of these three atoms in the compound
C : H : O
10 : 30 : 10
Divide the ratio by two to get simplest ratio
C : H : O
10/10 : 30/10 : 10/10
1 : 3 : 1
So for the empirical formula the simplest ratio of carbon to hydrogen to oxygen is 1:3:1
So the empirical formula will be
Empirical formula of C₁₀H₃₀O₁₀ = CH₃O
Now
To find the empirical formula mass in g/mol
Formula mass:
Formula mass is the total sum of the atomic masses of all the atoms present in a formula unit.
**Note:
if we represent the molar mass of the empirical formula for one mol in grams then it is written as g/mol
So,
As the empirical formula of C₁₀H₃₀O₁₀ is CH₃O
Then Its empirical formula mass will be
CH₃O
Atomic Mass of C = 12
Atomic Mass of H = 3
Atomic Mass of O = 16
Total Molar mass of CH₃O
CH₃O = 12 + 3(1) + 16
CH₃O = 12 + 3 + 16
CH₃O = 31 g/mol
<u>Given:</u>
Initial amount of carbon, A₀ = 16 g
Decay model = 16exp(-0.000121t)
t = 90769076 years
<u>To determine:</u>
the amount of C-14 after 90769076 years
<u>Explanation:</u>
The radioactive decay model can be expressed as:
A = A₀exp(-kt)
where A = concentration of the radioactive species after time t
A₀ = initial concentration
k = decay constant
Based on the given data :
A = 16 * exp(-0.000121*90769076) = 16(0) = 0
Ans: Based on the decay model there will be no C-14 left after 90769076 years
An Insulator.
Insulators prevent flow of electric current. Example is wood, clay.
Example of solid - solid homogeneous mixture is copper metal - silver metal like coins and alloys.
Homogeneous mixture is a mixture in which one of the substances often changes in form as in a solution of sugar in water. It contains variable proportions. Solution can contain two substances, three substances or more, in a single physical state. The component of a solution that is present in greatest quantity is usually called the solvent and all other components are called solutes.
Note that it says oxygen "gas"
So you need the atomic mass of oxygen gas
Look at your periodic table, you'll see 15.9994 under oxygen
Oxygen gas has a formula of O2 therefore,
(15.9994) times 2= Oxygen gas atomic mass=31.9988
Mol= Mass/Atomic Mass
=62.3 g/ 31.9988 g/mol = 1.95 mol
now look at the ratio of C2H6 and O2, notice there is an invisible number beside each of them, at that "invisible number" is =1
1 C2H6 + 1 O2 -> products
this means that for 1 mol of C2H6, 1 mol of O2 has to react with it
Thus as we have 1.95 moles of O2, we need 1.95 moles of C2H6
