What do they mean by accepted molar mass of butane in question #7? I've calculated the molar mass of butane gas and found that it was 0.574 g/mol because I found the mass of the butane (0.1027 g) and divided it by the number of moles of butane (0.179 mol). **Educational Resource: Experimental Determination of Molar Mass of Butane****Data Table:**A table is presented with the following columns:1. **Trial Number**: (1, 2, 3)2. **Lighter Number**: (Blank rows)3. **Initial Mass of Lighter** (g): - 13.5099 - Blank - Blank 4. **Final Mass of Lighter** (g): - 12.0482 - Blank - Blank - (Note: "we’ll get this after they have dried overnight")5. **Temperature** (°C) (Reads from the thermometer): - 22.4 - Blank - Blank 6. **Volume of Butane** (mL) (Reads from eudiometer tube): - 448 - 2 - Blank 7. **Barometric Pressure** (inHg) (Reads from barometer): - 29.85 - Blank - Blank 8. **Vapor Pressure of Water** (Reads from chart of vapor pressures): - 20.45 - Blank - Blank **Analysis:**1. **Convert** the barometric pressure from inHg to mmHg. (1 in = 25.4 mm).2. **Use Dalton’s Law** to calculate the pressure of the butane alone.3. Using the **Ideal Gas Law**, calculate the moles of butane for each trial using the formula: \[ PV = nRT \]4. **Calculate** the mass of butane collected for each trial.5. **Calculate** the molar mass of butane for each trial using: \[ \text{Molar Mass} = \frac{\text{Mass}}{\# \text{ of moles}} \]6. **Calculate** the average molar mass of butane.7. **Determine** the formula of butane and calculate the percent error in your experimentally determined molar mass.8. If some bubbles of butane escape and are not collected in the eudiometer, how will the experimentally determined molar mass be affected?9. If an air bubble is left in the eudiometer before the gas is collected, how will the experimentally determined molar mass be

Introduction: The accepted molar mass is nothing but the actual molar mass calculated from the molecular formula of the substance. The experimental molar mass can be different from the accepted molar mass. Calculation part: Given, Initial mass of the lighter = 13.3509 g, and the final mass of the lighter = 13.2482 g. Therefore the actual mass of the butane = (13.3509 - 13.2482) g = 0.1027 g. Temperature (T) = 22.4oC = 22.4 + 273 K = 295.4 K Volume of butane (V) = 4482 ml = 4.482 L Barometric pressure = 29.85 mmHg, Vapor pressure of water = 20.45 mmHg Therefore, the pressure of butane (P) = 29.85 - 20.45 mmHg = 9.4 mmHg = 9.4 x 0.001316 atm = 0.0124 atm Now, from PV = nRT equation, n (mole) = PVRT = 0.0124 (atm)×4.482 (L)0.082 (L.atm mol-1K-1)×295.4(K) (Universal gas constant, R= 0.082 L.atm mol-1K-1) = 0.0023 moles Now, the experimental molar mass = massmole= 0.1027 g0.0023 mole = 44.65 gmol-1 Again the accepted molar mass from the molecular formula of the butane (C4H10) = (4 x 12.011) + (1.00789 x 10) gmol-1 = 58.12 gmol-1 (Answer) Therefore, percent error with respect to the accepted molar mass = experimental - actualactualx100% =44.65 - 58.1258.12 ×100% = 23.2% (Answer)