Caffeine, a stimulant found in coffee, tea, and certain soft drinks, contains c, h, o, and n. combustion

Caffeine, a stimulant found in coffee, tea, and certain soft drinks, contains c, h, o, and n. combustion of 1.000 mg of caffeine produces 1.813 mg co2, 0.4639 mg h2o, and 0.2885 mg n2. estimate the molar mass of caffeine, which lies between 150 and 200 g/mol.

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  1. Assuming 1 mole of caffeine, therefore the mass is 194.19 grams.

    C = 0.4948 * 194.19 grams * (1 mol / 12 grams) = 8

    H = 0.0519 * 194.19 grams * (1 mol / 1 grams) = 10

    N = 0.2885 * 194.19 grams * (1 mol / 14 grams) = 4

    O = 0.1648 * 194.19 grams * (1 mol / 16 grams) = 2

     

    So the molecular formula is:

    C8H10N4O2

  2. 1.) 1 mole. 2.) molar mass of C8H10N4O2 = 8x molar mass of C + 10x molar mass of H + 4x molar mass of N + 2x molar mass of O = 8x12 + 10x1 + 4x14 + 2x16 = 194gm/mol. 3.) Balanced equation is 2AlCl3 (s) + Ca3N2 (s) > 2AlN + 3CaCl2.

  3. Molar Mass=193.9738 g/mol

    Molar Mass≅194g/mol

    Explanation:

    Consider the formula:

    m=ΔT/K_{f}

    where:

    ΔT is freezing point depression

    K_{f} is freezing point depression constant

    m is the morality=(moles of solute/kg of solvent)

    Now:

    [tex]m=\frac{3.07}{39.7}[/tex]

    [tex]m=0.07733 moles of solute/kg of solvent[/tex]

    Now:

    0.07733 (moles of solute/kg of solvent) *0.010 kg of solvent

    [tex]7.733*10^{-4}[/tex] moles of solute(Caffeine)

    Molar mass = Mass of solute/moles of solute

    Molar Mass=[tex]\frac{150*10^-3}{7.733*10^{-4} }[/tex]

    Molar Mass=193.9738 g/mol

    Molar Mass≅194g/mol

  4. 59.0764

    Explanation:

    ElementSymbolAtomic weightAtoms

    Calcium   Ca             40.078              1        

    Fluorine     F             18.9984032             1  

    you just multiply the atomic weight to the atom and then add :))

  5. 194 g/mol

    Explanation:

    1) Content of C:

    All the C atoms in the 1.000 mg of caffeine will be found in the 1.813 mg of CO₂.

    Mass of C in 1.813 mg of CO₂

           Since the atomic mass of C is 12.01 g/mol and the molar of of CO₂ is 44.01, there are 12 mg of C in 44 mg of CO₂ and you can set the proporton:

           12.01 mg C / 44.01 mg CO₂ = x / 1.813 g CO₂

            ⇒ x = 1.813 × 12.01 / 44.01 g of C = 0.49475 mg of C

    Number of moles of C

          number of moles = mass in g / atomic mass = 0.49475×10⁻³ g / 12.01 g/mol = 4.1195×10⁻⁵ moles = 0.041195 milimol

    2) Content of H

    All the H atoms in the 1.000 mg of caffeine will be found in the 0.4639 mg of H₂O

    Mass of H in 0.4639 mg of H₂O

           Since the atomic mass of H is 1.008 g/mol and the molar of of H₂O is 18.015 g/mol, there are 2×1.008 mg of H in 18.015 mg of H₂O and you can set the proporton:

           2×1.008 mg H / 18.015 mg H₂O = x / 0.4639 mg H₂O

            ⇒ x = 0.4639 mg H₂O × 2 × 1.008 mg H / 18.015 mg H₂O = 0.051913 mg H

    Number of moles of H

          number of moles = mass in g / atomic mass = 0.051913 ×10⁻³ g / 1.008 g/mol = 5.1501× 10⁻⁵ moles = 0.051501 milimol

    3) Content of N

    All the N atoms in the 1.000 mg of caffeine will be found in the 0.2885 mg of N₂

    Mass of N in 0.2885 mg of N₂ is 0.2885 mgNumber of moles of N

          number of moles = mass in g / atomic mass = 0.2885 ×10⁻³ g / 14.007 g/mol = 2.0597× 10⁻⁵ moles = 0.020597 milimol

    4) Content of O

    The mass of O is calculated by difference:

    Mass of O = mass of sample - mass of C - mass of H - mass of N

           Mass of O = 1.000 mg - 0.49475 mg C - 0.051913 mg H - 0.2885 mg N

         Mass of O = 0.1648 mg

    Moles of O =  0.1648 × 10 ⁻³ g / 15.999 g/mol = 1.0303×10⁻⁵ mol = 0.01030 milimol

    5) Ratios

    Divide every number of mililmoles by the smallest number of milimoles:

    C:  0.041195 / 0.01030 = 4H: 0.051501 / 0.01030 = 5N: 0.020597 / 0.01030 = 2O: 0.01030 / 0.01030 = 1C: 4H: 5N: 2O: 1

    6) Empirical formula:

    C₄H₅N₂O₁

    7) Calculate the approximate mass of the empirical formula:

    4 × 12 + 5 × 1 + 2 × 14 + 1 × 16 =  97 g/mol

    So, since that number is not between 150 and 200 g/mol, multiply by 2: 97 × 2 = 194, which is between 150 and 200.

    Thus, the estimate is 194 g/mol

  6. The formula mass of caffeine is 97 amu and molar mass of caffeine is 194 g/mol

    Explanation:

    Formula mass is defined as the sum of the mass of all the atoms each multiplied its atomic masses that are present in the empirical formula of a compound. It is expressed in amu.

    Molar mass is defined as the sum of the mass of all the atoms each multiplied its atomic masses that are present in the molecular formula of a compound. It is expressed in g/mol.

    Empirical formula is defined as the formula in which atoms in a compound are present in simplest whole number ratios.

    The molecular formula of caffeine is [tex]C_8H_{10}N_4O_2[/tex]

    Dividing each number of atoms by '2', we will get the empirical formula of caffeine. The empirical formula of caffeine is [tex]C_4H_5N_2O[/tex]

    We know that:

    Atomic mass of carbon = 12 amu

    Atomic mass of hydrogen = 1 amu

    Atomic mass of nitrogen = 14 amu

    Atomic mass of oxygen = 16 amu

    Formula mass of caffeine = [tex](4\times 12)+(5\times 1)+(2\times 14)+(1\times 16)]=97amu[/tex]

    Molar mass of caffeine = [tex](8\times 12)+(10\times 1)+(4\times 14)+(2\times 16)]=194g/mol[/tex]

    Hence, the formula mass of caffeine is 97 amu and molar mass of caffeine is 194 g/mol

  7. The molecular formula of Caffeine is C₈H₁₀N₄O₂
    the mass of 1 mol of caffeine - 194 g
    In 1 mol of caffeine, mass of C is - 12 x8 = 96 g
    Since 1 mol - 194 g, in 194 g of caffeine there's 96 g of Carbon
    To find the mass percentage of C 
    Mass percentage = Mass of Carbon / mass of caffeine  x 100% 
                                =  96 g / 194 g x 100 
                                = 49.5 %
    49.5 % of caffeine is composed of C.

  8. First establish that your going to measure using 1 mole 
    So now you have 194.19g caffeine 

    Carbon is 49.48% of the mass 
    so you have (49.48%)*(194.19)=96.0852 g carbon 
    to see how many carbons you have you do (mass/Molar mass) 
    (96.0852/12.0107) = 10 
    So you have 10 carbons 

    Do it for each one: 
    Hydrogen: 
    (5.19%)*(194.19) = 10.0785 g 
    (mass/molar mass) (10.0785/1.00794) = 10 

    Nitrogen: 
    (28.85%)(194.19) = 56.0238 g 
    (56.0238/14.00674) = 4 

    Oxygen: 
    (16.48%)(194.19) = 32.0025g 
    (32.0025/15.9994) = 2 

    Final formula: C8 H10 N4 O2

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