Ans: Have the same number of electron shells
The molecular formula of the compound is C12H15O3 hence the molar mass of the compound is 207 g/mol.
We need to obtain the number of moles of carbon, hydrogen and oxygen in the compound;
Carbon = 24.91 g/44g/mol × 1 mole of carbon = 0.566 moles
Mass of carbon = 0.566 moles × 12 g/mol = 6.792 g
Number of moles of hydrogen = 6.522 g/18 g/mol × 2 moles = 0.725 moles
Mass of hydrogen = 0.725 moles × 1 g/mol = 0.725 g
Mass of oxygen = 10 - (6.792 g + 0.725 g) = 2.483 g
Number of moles of oxygen = 2.483 g/16 g/mol = 0.155 moles
Now we must divide through by the lowest number of moles;
C - 0.566/0.155 H - 0.725/0.155 O - 0.155/0.155
C - 4 H - 5 O - 1
The simplest formula is C4H5O Recall that the molar mass of the compound lies between 150.0 and 220.0 g/mol
4(12) + 5(1) + 16 = 69
Hence; n = 3 and the molecular formula of the compound is C12H15O3
The molar mass of the compound is; 12(12) + 15(1) + 3(16) = 207 g/mol
Learn more: brainly.com/question/15180604
Answer:
%N = 25.94%
%O = 74.06%
Explanation:
Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅
We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.
m(N): 2 × 14.01 g = 28.02 g
Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅
We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.
m(O): 5 × 16.00 g = 80.00 g
Step 3: Calculate the mass of 1 mole of N₂O₅
We will sum the masses of N and O.
m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g
Step 4: Calculate the percent composition of N₂O₅
We will use the following expression.
%Element = m(Element)/m(Compound) × 100%
%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%
%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%
False , it will make it lighter
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