Answer;
C7H14O2
Solution;
Isobutyl contains , oxygen, carbon and hydrogen (total mass is 1.152 g)
Mass of carbon = 12/44 × 2.726 g
= 0.743455 g
Mass of Hydrogen = 2/18 × 1.116 g
= 0.124 g
Mass of oxygen = 1.152 - (0.7435 + 0.124)
= 0.2845 g
Moles of carbon ; 0.7435/12 = 0.06196 moles
Moles of hydrogen; 0.124/1 = 0.124 moles
Moles of oxygen; 0.2845/16 = 0.01778 moles
Ratios ; 0.06196/0.01778 ; 0.124/0.01778 : 0.01778/0.01778
= 3.5 : 7.0 : 1
To make them whole numbers ; we multiply the ratios by 2 to get;
(3.5 : 7.0 : 1 )2 = 7 : 14 : 2
Thus, the empirical formula of Isobutyl propionate is C7H14O2
V ( HCl ) = 45.00 mL in liters : 45.00 / 1000 => 0.045 L
M ( HCl ) = ?
V ( NaOH ) = 25.00 / 1000 => 0.025 L
M ( NaOH) = 0.2000 M
number of moles NaOH :
n = M x V = 0.2000 x 0.025 => 0.005 moles of NaOH
Mole ratio:
HCl + NaOH = NaCl + H2O
1 mole HCl ---------- 1 mole NaOH
? mole HCl ---------- 0.005 moles NaOH
moles HCl = 0.005 x 1 / 1
= 0.005 moles of HCl :
M ( HCl ) = n / V
M ( HCl ) = 0.005 / 0.045
= 0.1111 M
hope this helps!
Steps/Answers:
Jot down the metal name.
Write the metal name ending in -idle.
Combine cation, annion names,
Hope This Helps! Have A Nice Day!!
D. It gives the same results when experiments are repeated
The mass percent of hydrogen in CH₄O is 12.5%.
<h3>What is the mass percent?</h3>
Mass percent is the mass of the element divided by the mass of the compound or solute.
- Step 1: Calculate the mass of the compound.
mCH₄O = 1 mC + 4 mH + 1 mO = 1 (12.01 amu) + 4 (1.00 amu) + 1 (16.00 amu) = 32.01 amu
- Step 2: Calculate the mass of hydrogen in the compound.
mH in mCH₄O = 4 mH = 4 (1.00 amu) = 4.00 amu
- Step 3: Calculate the mass percent of hydrogen in the compound.
%H = (mH in mCH₄O / mCH₄O) × 100%
%H = 4.00 amu / 32.01 amu × 100% = 12.5%
The mass percent of hydrogen in CH₄O is 12.5%.
Learn more about mass percent here:brainly.com/question/4336659
