A compound contains 4.07% hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?
Answers
Answer:
180.156 g/mol
Formula: C₆H₁₂O₆
Density: 1.56 g/cm³
IUPAC ID: D-glucose
Explanation:
SRY FIR SE BUT TUNE RRPLY NHI DIYA XD ESA KYU KYU ?
# ITURSAAYUSH XD
Answer:
step 1
step 1Divide the given percentages of atoms with their molecular masses
step 1Divide the given percentages of atoms with their molecular massesH---4.07/1 =4.07
step 1Divide the given percentages of atoms with their molecular massesH---4.07/1 =4.07C---24.27/12 =2.02
step 1Divide the given percentages of atoms with their molecular massesH---4.07/1 =4.07C---24.27/12 =2.02Cl--71.65/35.5=2.01
step 1Divide the given percentages of atoms with their molecular massesH---4.07/1 =4.07C---24.27/12 =2.02Cl--71.65/35.5=2.01step 2
step 1Divide the given percentages of atoms with their molecular massesH---4.07/1 =4.07C---24.27/12 =2.02Cl--71.65/35.5=2.01step 2divide all values with the lowest value obtained.
step 1Divide the given percentages of atoms with their molecular massesH---4.07/1 =4.07C---24.27/12 =2.02Cl--71.65/35.5=2.01step 2divide all values with the lowest value obtained. H---4.07/2.01=2
step 1Divide the given percentages of atoms with their molecular massesH---4.07/1 =4.07C---24.27/12 =2.02Cl--71.65/35.5=2.01step 2divide all values with the lowest value obtained. H---4.07/2.01=2 C---2.02/2.01=1
step 1Divide the given percentages of atoms with their molecular massesH---4.07/1 =4.07C---24.27/12 =2.02Cl--71.65/35.5=2.01step 2divide all values with the lowest value obtained. H---4.07/2.01=2 C---2.02/2.01=1 Cl---2.01/2.01=1
step 1Divide the given percentages of atoms with their molecular massesH---4.07/1 =4.07C---24.27/12 =2.02Cl--71.65/35.5=2.01step 2divide all values with the lowest value obtained. H---4.07/2.01=2 C---2.02/2.01=1 Cl---2.01/2.01=1therefore the empirical formula is
step 1Divide the given percentages of atoms with their molecular massesH---4.07/1 =4.07C---24.27/12 =2.02Cl--71.65/35.5=2.01step 2divide all values with the lowest value obtained. H---4.07/2.01=2 C---2.02/2.01=1 Cl---2.01/2.01=1therefore the empirical formula isCH 2
step 1Divide the given percentages of atoms with their molecular massesH---4.07/1 =4.07C---24.27/12 =2.02Cl--71.65/35.5=2.01step 2divide all values with the lowest value obtained. H---4.07/2.01=2 C---2.02/2.01=1 Cl---2.01/2.01=1therefore the empirical formula isCH 2
step 1Divide the given percentages of atoms with their molecular massesH---4.07/1 =4.07C---24.27/12 =2.02Cl--71.65/35.5=2.01step 2divide all values with the lowest value obtained. H---4.07/2.01=2 C---2.02/2.01=1 Cl---2.01/2.01=1therefore the empirical formula isCH 2 Cl
step 1Divide the given percentages of atoms with their molecular massesH---4.07/1 =4.07C---24.27/12 =2.02Cl--71.65/35.5=2.01step 2divide all values with the lowest value obtained. H---4.07/2.01=2 C---2.02/2.01=1 Cl---2.01/2.01=1therefore the empirical formula isCH 2 ClWEIGHT OF EMPIRICAL FORMULA=12+2+35.5=49.5
step 1Divide the given percentages of atoms with their molecular massesH---4.07/1 =4.07C---24.27/12 =2.02Cl--71.65/35.5=2.01step 2divide all values with the lowest value obtained. H---4.07/2.01=2 C---2.02/2.01=1 Cl---2.01/2.01=1therefore the empirical formula isCH 2 ClWEIGHT OF EMPIRICAL FORMULA=12+2+35.5=49.5Given molecular weight =98.96 which is double of empirical weight .therefore molecular formula is C ²H ⁴Cl² .