Your data looks like this:

df
   Compound      conc    Count conc.pred
1         A 3.720e-03 10200000  565.1263
2         A 1.368e-02 10300000  559.3499
3         A 5.004e-02 10200000  565.1263
4         A 1.836e-01 10200000  565.1263
5         A 6.732e-01  9927000  580.8959
6         A 2.472e+00  9400000  611.3376
7         A 9.048e+00  8464040  665.4025
8         A 3.324e+01  6513840  778.0542
9         A 1.212e+02  4349320  903.0858
10        A 4.464e+02  3071480  976.8991
11        A 1.632e+03  5989320  808.3526
12        A 6.000e+03  8444560  666.5278

What are you aiming to model? You cannot do anything with the compound column because it only has one value.

The only thing that we can do here is build a linear regression model between conc and Count:

model <- lm(conc ~ Count, data = df)
summary(model)

Call:
lm(formula = conc ~ Count, data = df)

Residuals:
   Min     1Q Median     3Q    Max 
-781.9 -620.7 -565.1 -552.1 5333.5 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)
(Intercept)  1.154e+03  1.836e+03   0.629    0.544
Count       -5.776e-05  2.176e-04  -0.265    0.796

Residual standard error: 1816 on 10 degrees of freedom
Multiple R-squared:  0.006999,  Adjusted R-squared:  -0.0923 
F-statistic: 0.07049 on 1 and 10 DF,  p-value: 0.796

It looks like there is no statistically significant relationship between these.

We can assess R^2 shrinkage using 3-fold cross-validation:

library(bootstrap)
theta.fit <- function(x, y) { lsfit(x,y) }
theta.predict <- function(fit, x) { cbind(1,x) %*% fit$coef }
x <- as.matrix(df[,which(colnames(df) %in% names(summary(model)$coefficients[,1]))])
y <- as.matrix(df[,c("conc")]) 
results <- crossval(x, y, theta.fit, theta.predict, ngroup = 3)

# Raw R^2 
cor(y, model$fitted.values)**2
            [,1]
[1,] 0.006999459

# cross-validated R^2
cor(y, results$cv.fit)**2
           [,1]
[1,] 0.07150013

It does not look like a good model, in fairness.

We can also plot the model-predicted conc values versus the real ones:

require(ggplot2)
df$conc.pred = predict(model)
ggplot(df, aes(x = conc, y = conc.pred)) +
  geom_point() +
  geom_smooth(method='lm',formula=y~x) +
  stat_smooth(method="lm", fullrange=TRUE) +
  ggtitle("My linear model")

The confidence intervals (grey shade) are extraordinary.

Finally, we can plot the ROC curve:

require(pROC)
roc <- roc(df$conc, fitted(model), smooth=FALSE)
plot.roc(
  roc,
  grid=TRUE,
  grid.lwd=2,
  col="royalblue",
  main="My linear model")
text(0.3, 0.45,
  col="royalblue",
  paste("AUC (L95, AUC, U95):",
    paste(round(ci.auc(roc)[1:3], 3), sep=" ", collapse=","), sep=" "),
  cex=1.0)

Hopefully this can 'foment' some ideas for you.

If you have other compounds in your data and want to model them as the y variables in the model, then you can switch to using a generalised linear model (glm()) in place of the linear model (lm()).

Kevin