Answer:
Explanation:
1. find the molar mass (amu) of each element and add them to get the whole molar mass.
2. divide the 1 element molar mass with the whole molar mass
3. multiple by 100 and that gives you the % composition.
56-57: NaCl
1. Na(22.99amu) + Cl (35.453amu)=58.443
2(Na): [tex]\frac{22.99}{58.443}[/tex] = .393
2(Cl): [tex]\frac{35.453}{58.443}[/tex]= .607
3(Na): .393 * 100=39.3%
3(Cl): .607 * 100= 60.7%
58-60 [tex]K_{2} CO_{3}[/tex]
1. K: (39.098)(2)=78.196
_ C: (12.011)(1)= 12.011
_O: (15.99)(3) = 47.997
78.196+12.011+47.997= 138.204
2:K: [tex]\frac{78.196}{138.204}[/tex]= .566 Step 3: (.566)(100)= 56.6%
2: C: [tex]\frac{12.011}{138.204}[/tex]= .087 Step 3: (.087)(100)= 8.7%
2: O: [tex]\frac{47.997}{138.204}[/tex]= .347 Step 3: (.347)(100) = 34.7%
61-62 [tex]Fe_{3} O_{4}[/tex]
1. Fe (55.845)(3)= 167.535
_ O (15.999)(4) = 63.996
167.535+63.996=231.531
2: Fe: [tex]\frac{167.535}{231.531}[/tex]= .724 Step 3: (.724)(100)= 72.4%
2: O : [tex]\frac{63.996}{231.531}[/tex]= .276 Step 3: (.276)(100) = 27.6%
63-65 [tex]C_{3}H_{5}(OH)_{3}[/tex]
1.
C(12.011*3)=36.033
H(1.008*5)=5.04 + (1.008*3)=3.024 so its 8.064
O(15.999*3)=47.997
add them: 92.094
2: C: [tex]\frac{36.033}{92.094}[/tex]= .391 Step 3: (.391)(100) = 39.1%
2: H: [tex]\frac{8.064}{92.094}[/tex]= .088 step 3: (.088)(100) = 8.8%
2: O: [tex]\frac{47.997}{92.094}[/tex] = .521 step 3: (.521)(100) = 52.1%
