Answer:
0.001 M OH-
Explanation:
[OH-] = 10^-pOH, so
pOH + pH = 14 and 14 - pH = pOH
14 - 11 = 3
[OH⁻] = 10⁻³ ; [OH-] = 0.001 M OH-
4 carbon electrons. Hope this helps have a nice day
Answer:
Empirical formula is C₉H₁₅O
Molecular formula = C₈₁H₁₃₅O
₉
Explanation:
Percentage of carbon = 77.60%
Percentage of oxygen = 11.45%
Percentage of hydrogen = 10.95%
Molecular weight = 1253 g/mol
Molecular formula = ?
Empirical formula = ?
Solution:
Number of gram atoms of C = 77.60 g /12g/mol =6.5
Number of gram atoms of O = 11.45 g / 16 g/mol = 0.72
Number of gram atoms of H = 10.95 g / 1.008 g/mol= 10.9
Atomic ratio:
C : H : O
6.5/0.72 : 10.9/0.72 : 0.72/0.72
9 : 15 : 1
C : H : O = 9 : 15 : 1
Empirical formula is C₉H₁₅O
Molecular formula:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
n = 1253 / 139
n = 9
Molecular formula = n (empirical formula)
Molecular formula = 9 (C₉H₁₅O
)
Molecular formula = C₈₁H₁₃₅O
₉
Answer:
In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.
Explanation:
tally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:
\displaystyle \%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%%H=
mass compound
mass H
×100%
\displaystyle \%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%%C=
mass compound
mass C
×100%
If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:
\displaystyle \%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%%H=
10.0g compound
2.5g H
×100%=25%
\displaystyle \%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%%C=
10.0g compound
7.5g C
×100%=75%
